RD Sharma Solutions Class 10 Trigonometric Ratios Exercise 5.1

RD Sharma Solutions Class 10 Chapter 5 Exercise 5.1

RD Sharma Class 10 Solutions Chapter 5 Ex 5.1 PDF Download

Exercise: 5.1

 

 1.) Find the value of Trigonometric ratios in each of the following provided one of the six trigonometric ratios are given.

Sol.

(i) \(\sin A = \frac{2}{3}\) 

Given:

\(\sin A = \frac{2}{3}\) ….. (1)

By definition,

\(\sin A = \frac{2}{3}\)  =  \(\frac{Perpendicular}{Hypotenuse}\) ….(2)

By Comparing (1) and (2)

We get,

Perpendicular side = 2 and

Hypotenuse = 3

1

Therefore, by Pythagoras theorem,

AC2 = AB2 + BC2

Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side

(AB)

Therefore,

32 = AB2 + 22

AB2 = 32 – 22

AB2 = 9 – 4

AB2 = 5

AB= \(\sqrt{5}\)

Hence, Base = \(\sqrt{5}\)

Now, \(\cos A = \frac{Base}{Hypotenuse}\)

\(\cos A = \frac{\sqrt{5}}{3}\)

Now, cosec A = \(\frac{1}{sin A}\)

Therefore,

cosec A = \(\frac{Hypotenuse}{Perependicular}\)

cosec A =\(\frac{3}{2}\)

Now, sec A = \(\frac{Hypotenuse}{Base}\)

Therefore,

sec A = \(\frac{3}{\sqrt{5}}\)

Now, tan A = \(\frac{Perependicular}{Base}\)

tan A =  \(\frac{2}{\sqrt{5}}\)

Now, cot A = \(\frac{Base}{Perpendicular}\)

Therefore,

cot A = \(\frac{\sqrt{5}}{2}\)

 

(ii) \(\cos A= \frac{4}{5}\)

Given:    \(\cos A= \frac{4}{5}\) …. (1)

By Definition,

\(\cos A = \frac{Base}{Hypotenuse}\) …. (2)

By comparing (1) and (2)

We get,

Base =4 and

Hypotenuse = 5

1

Therefore,

By Pythagoras theorem,

AC2 = AB2 + BC2

Substituting the value of base (AB) and hypotenuse (AC) and get the perpendicular side

(BC)

52 = 42+ BC2

BC2 = 52 – 42

BC2= 25 – 16

BC2 = 9

BC= 3

Hence, Perpendicular side = 3

Now,

\(\sin A = \frac{2}{3}\)  =  \(\frac{Perpendicular}{Hypotenuse}\)

Therefore,

\(\sin A= \frac{3}{5}\)

Now, cosec A = \(\frac{1}{sinA}\)

Therefore,

cosec A=\(\frac{1}{sinA}\)

Therefore,

cosec A =\(\frac{Hypotenuse}{Perependicular}\)

cosec A = \(\frac{5}{3}\)

Now, sec A =\(\frac{1}{cos A}\)

Therefore,

sec A =\(\frac{Hypotenuse}{Base}\)

sec A = \(\frac{5}{4}\)

Now, tan A = \(\frac{Perpendicular}{Base}\)

Therefore,

tan A =\(\frac{3}{4}\)

Now, cot A = \(\frac{1}{tan A}\)

Therefore,

cot A = \(\frac{Base}{Perpendicular}\)

cot A = \(\frac{4}{3}\)

 

(iii) \(\tan \Theta = \frac{11}{1}\)

Given:   \(\tan \Theta = \frac{11}{1}\) …. (1)

By definition,

\(\tan \Theta = \frac{Perpendicular}{Base}\) …. (2)

By Comparing (1) and (2)

We get,

Base= 1 and

Perpendicular side= 5

2

Therefore,

By Pythagoras theorem,

AC2 = AB2 + BC2

Substituting the value of base side (AB) and perpendicular side (BC) and get hypotenuse(AC)

AC2 = 12 + 112

AC2 = 1 + 121

AC2= 122

AC= \(\sqrt{122}\)

Now, \(\sin \Theta = \frac{Perpendicular}{Hypotenuse}\)

Therefore,

\(\sin \Theta = \frac{11}{\sqrt{122}}\)

Now, cosec\(\Theta =\frac{1}{sin\Theta }\)

cosec \(\Theta =\frac{\sqrt{122}}{11}\)

Now, \(\cos \Theta = \frac{Base}{Hypotenuse}\)

Therefore,

\(\cos \Theta = \frac{1}{\sqrt{122}}\)

Now, \(\sec \Theta =\frac{1}{cos\Theta }\)

Therefore,

\(\sec \Theta =\frac{Hypotenuse}{Base }\)

\(\sec \Theta =\frac{\sqrt{122}}{1}\)

\(\sec \Theta =\sqrt{122}\)

Now, \(\cot \Theta = \frac{1}{tan\Theta }\)

Therefore,

\(\cot \Theta = \frac{Base}{Perpendicular}\)

\(\cot \Theta = \frac{1}{11}\)

 

 

(iv) \(\sin \Theta = \frac{11}{15}\)

Given:   \(\sin \Theta = \frac{11}{15}\) …. (1)

By definition,

\(\sin \Theta = \frac{Perpendicular}{Hypotenuse}\) …. (2)

By Comparing (1) and (2)

We get,

Perpendicular Side = 11 and

Hypotenuse= 15

Therefore,

By Pythagoras theorem,

AC2 = AB2 + BC2

1

Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)

152 = AB2 +112

AB2 = 152 – 112

                        AB2= 225 – 121

AB2 = 104

AB = \(\sqrt{104}\)

AB= \(\sqrt{2\times 2\times 2\times 13}\)

AB= 2\(\sqrt{2\times 13}\)

AB= 2\(\sqrt{26}\)

Hence, Base = 2\(\sqrt{26}\)

Now, \(\cos \Theta = \frac{Base}{Hypotenuse}\)

Therefore,

\(\cos \Theta = \frac{2\sqrt{26}}{15}\)

Now, cosec[\(\Theta = \frac{1}{\sin \Theta }\)

Therefore,

cosec\(\Theta = \frac{Hypotenuse}{Perpendicular}\)

cosec\(\Theta = \frac{15}{11}\)

Now, sec\(\Theta = \frac{Hypotenuse}{Base }\)

Therefore,

sec\(\Theta = \frac{15}{2\sqrt{26}}\)

Now, \(\tan \Theta =\frac{Perpendicular}{Base}\)

Therefore,

\(\tan \Theta =\frac{11}{2\sqrt{26}}\)

Now, \(\cot \Theta =\frac{Base}{Perpendicular}\)

Therefore,

\(\cot \Theta =\frac{2\sqrt{26}}{11}\)

 

(v) \(\tan \alpha = \frac{5}{12}\)

Given:   \(\tan \alpha = \frac{5}{12}\) …. (1)

By definition,

\(\tan \alpha = \frac{Perpendicular}{Base}\) …. (2)

By comparing (1) and (2)

We get,

Base= 12 and

Perpendicular side = 5

1

Therefore,

By Pythagoras theorem,

AC2 = AB2 + BC2

Substituting the value of base side (AB) and the perpendicular side (BC) and gte hypotenuse (AC)

AC2 = 122 + 52

AC2 = 144 + 25

AC2= 169

AC= 13

Hence Hypotenuse = 13

Now, \(\sin \alpha = \frac{Perpendicular}{Hypotenuse}\)

Therefore,

\(\sin \alpha = \frac{5}{13}\)

Now, cosec\(\alpha = \frac{Hypotenuse}{Perpendicular}\)

cosec\(\alpha = \frac{13}{5}\)

Now, \(\cos \alpha = \frac{Base}{Hypotenuse}\)

Therefore,

\(\cos \alpha = \frac{12}{13}\)

Now, \(\sec \alpha =\frac{1}{cos\alpha }\)

Therefore,

\(\cot \alpha =\frac{Base}{Perpendicular}\)

\(\cot \alpha =\frac{12}{5}\)

 

(vi) \(\sin \Theta =\frac{\sqrt{3}}{2}\)

Given:   \(\sin \Theta =\frac{\sqrt{3}}{2}\) …. (1)

By definition,

\(\sin \Theta =\frac{Perpendicular}{Hypotenuse}\) ….(2)

By comparing (1) and (2)

We get,

Perpendicular Side = \(\sqrt{3}\)

Hypotenuse = 2

1

Therefore,

By Pythagoras theorem,

AC2 = AB2 + BC2

Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)

22 = AB2 + (\(\sqrt{3}\))2

AB2 = 22 – (\(\sqrt{3}\))2

AB2 = 4 – 3

AB2 = 1

AB= 1

Hence Base = 1

Now, \(\cos \Theta = \frac{Base}{Hypotenuse}\)

Therefore,

\(\cos \Theta = \frac{1}{2}\)

Now, cosec\(\Theta =\frac{1}{\sin \Theta }\)

Therefore,

cosec\(\Theta =\frac{Hypotenuse}{Perpendicualar }\)

cosec\(\Theta =\frac{2}{\sqrt{3}}\)

Now, \(\sec \Theta = \frac{Hypotenuse}{Base}\)

Therefore,

\(\sec \Theta = \frac{2}{1}\)

Now, \(\tan \Theta = \frac{Perpendicular}{Base}\)

Therefore,

\(\tan \Theta = \frac{\sqrt{3}}{1}\)

Now, \(\cot \Theta = \frac{Base}{Perpendicular}\)

Therefore,

\(\cot \Theta = \frac{1}{\sqrt{3}}\)

 

(vii) \(\cos \Theta = \frac{7}{25}\)

Given:   \(\cos \Theta = \frac{7}{25}\) …. (1)

By definition,

\(\cos \Theta = \frac{Base}{Hypotenuse}\)

By comparing (1) and (2)

We get,

Base = 7 and

Hypotenuse = 25

1

Therefore

By Pythagoras theorem,

AC2= AB2 + BC2

Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

252 = 72 +BC2

BC2 = 252 – 72

BC2 = 625 – 49

BC = 576

BC= \(\sqrt{576}\)

BC= 24

Hence, Perpendicular side = 24

Now, \(\sin \Theta = \frac{perpendicular}{Hypotenuse}\)

Therefore,

\(\sin \Theta = \frac{24}{25}\)

Now, cosec\(\Theta = \frac{1}{\sin \Theta }\)

Therefore,

cosec\(\Theta = \frac{Hypotenuse}{Perpendicualar}\)

cosec\(\Theta = \frac{25}{24}\)

Now, \(\sec \Theta = \frac{1}{\cos \Theta }\)

Therefore,

\(\sec \Theta = \frac{Hypotenuse}{Base }\)

\(\sec \Theta = \frac{25}{7}\)

Now, \(\tan \Theta = \frac{Perpendicular}{Base}\)

Therefore,

\(\tan \Theta = \frac{24}{7}\)

Now, \(\cot \Theta = \frac{1}{\tan \Theta }\)

Therefore,

\(\cot \Theta = \frac{Base}{Perpendicular }\)

\(\cot \Theta = \frac{7}{24}\)

 

                (viii) \(\tan \Theta = \frac{8}{15}\)

Given:   \(\tan \Theta = \frac{8}{15}\) …. (1)

By definition,

\(\tan \Theta = \frac{Perpendicular}{Base}\) …. (2)

By comparing (1) and (2)

We get,

Base= 15 and

Perpendicular side = 8

1

Therefore,

By Pythagoras theorem,

AC2= 152 + 82

AC2= 225 + 64

AC2 = 289

AC = \(\sqrt{289}\)

AC= 17

Hence, Hypotenuse = 17

Now, \(\sin \Theta =\frac{Perpendicular}{Hypotenuse}\)

Therefore,

\(\sin \Theta =\frac{8}{17}\)

Now, cosec\(\Theta =\frac{1}{\sin \Theta }\)

Therefore,

cosec\(\Theta = \frac{Hypotenuse}{Perpendicular}\)

\(\Theta = \frac{17}{8}\)

Now, \(\cos \Theta = \frac{Base}{Hypotenuse}\)

Therefore,

\(\cos \Theta = \frac{15}{17}\)

Now, \(\sec \Theta = \frac{1}{\cos \Theta }\)

Therefore,

\(\sec \Theta = \frac{Hypotenuse}{Base}\)

\(\sec \Theta = \frac{17}{15}\)

Now, \(\cot \Theta = \frac{1}{\tan \Theta }\)

Therefore,

\(\cot \Theta = \frac{Base}{Perpendicular}\)

\(\cot \Theta = \frac{15}{8}\)

 

(ix) \(\cot \Theta = \frac{12}{5}\)

Given:   \(\cot \Theta = \frac{12}{5}\) …. (1)

By definition,

\(\cot \Theta = \frac{1}{\tan \Theta }\)

\(\cot \Theta = \frac{Base}{Perpendicular}\) …. (2)

By comparing (1) and (2)

We get,

Base = 12 and

Perpendicular side = 5

1

Therefore,

By Pythagoras theorem,

AC2= AB2 + BC2

Substituting the value of base side (AB) and perpendicular side(BC) and get the hypotenuse (AC)

AC2 = 122 + 52

AC2= 144 + 25

AC2 = 169

AC = \(\sqrt{169}\)

AC = 13

Hence, Hypotenuse = 13

Now, \(\sin \Theta = \frac{Perpendicular}{Hypotenuse}\)

Therefore,

\(\sin \Theta = \frac{5}{13}\)

Now, cosec\(\Theta = \frac{1}{\sin \Theta }\)

Therefore,

cosec\(\Theta = \frac{Hypotenuse }{Perpendicular }\)

cosec\(\Theta = \frac{13 }{5}\)

Now, \(\cos \Theta = \frac{Base}{Hypotenuse}\)

Therefore,

\(\cos \Theta = \frac{12}{13}\)

Now, \(\sec \Theta = \frac{1}{cos\Theta }\)

Therefore,

\(\sec \Theta = \frac{Hypotenuse}{Base }\)

\(\sec \Theta = \frac{13}{12}\)

Now, \(\tan \Theta = \frac{1}{\cot \Theta }\)

Therefore,

\(\tan \Theta = \frac{Perpendicular}{Base}\)

\(\tan \Theta = \frac{5}{12}\)

 

                (x) \(\sec \Theta = \frac{13}{5}\)

Given:   \(\sec \Theta = \frac{13}{5}\)… (1)

By definition,

\(\sec \Theta = \frac{1}{\cos \Theta }\) …. (2)

By comparing (1) and (2)

We get,

Base=5

Hypotenuse = 13

1

Therefore,

By Pythagoras theorem,

Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

132 = 52 + BC2

BC2 = 132 – 52

BC2=169 – 25

BC2= 144

BC= \(\sqrt{144}\)

BC = 12

Hence, Perpendicular side = 12

Now, \(\sin \Theta = \frac{Perpendicular}{Hypotenuse}\)

Therefore,

\(\sin \Theta = \frac{12}{13}\)

Now, cosec \(\Theta = \frac{1}{\sin \Theta }\)

Therefore,

cosec\(\Theta = \frac{Hypotenuse}{Perpendicular}\)

cosec\(\Theta = \frac{13}{12}\)

Now, \(\cos \Theta = \frac{1}{\sec \Theta }\)

Therefore,

\(\cos \Theta = \frac{Base}{Hypotenuse }\)

\(\cos \Theta = \frac{5}{13}\)

Now, \(\tan \Theta = \frac{Perpendicular}{Base}\)

Therefore,

\(\tan \Theta = \frac{12}{5}\)

Now, \(\cot \Theta = \frac{1}{\tan \Theta }\)

Therefore,

\(\cot \Theta = \frac{Base}{Perpendicular }\)

\(\cot \Theta = \frac{5}{12 }\)

 

               (xi) cosec\(\Theta= \sqrt{ 10 }\)

Given:   cosec\(\Theta = \frac{\sqrt{10}}{1}\)… (1)

By definition

cosec\(\Theta= \frac{1}{\sin \Theta }\) ….(2)

\(\Theta = \frac{Hypotenuse}{Perpendicular}\)

By comparing (1) and(2)

We get,

Perpendicular side= 1 and

Hypotenuse = \(\sqrt{10}\)

1

Therefore,

By Pythagoras theorem,

AC2 = AB2 + BC2

Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)

(\(\sqrt{10}\))2 = AB2 + 12

AB2= (\(\sqrt{10}\))2 – 12

AB2= 10 – 1

AB = \(\sqrt{9}\)

AB = 3

Hence, Base side = 3

Now, \(\sin \Theta = \frac{Perpendicular}{Hypotenuse}\)

Therefore,

\(\sin \Theta = \frac{1}{\sqrt{10}}\)

Now, \(\cos \Theta = \frac{Base}{Hypotenuse}\)

Therefore,

\(\cos \Theta = \frac{3}{\sqrt{10}}\)

Now, \(\sec \Theta = \frac{1}{\cos \Theta }\)

Therefore,

\(\sec \Theta = \frac{Hypotenuse}{Base}\)

\(\sec \Theta = \frac{\sqrt{10}}{3}\)

Now, \(\tan \Theta = \frac{Perpendicular}{Base}\)

Therefore,

\(\tan \Theta = \frac{1}{3}\)

Now, \(\cot \Theta = \frac{1}{\tan \Theta }\)

\(\cot \Theta = \frac{3}{1}\)

\(\cot \Theta = 3\)

 

               (xii) \(\cos \Theta \frac{12}{15}\)

Given:   \(\cos \Theta \frac{12}{15}\) …. (1)

By definition,

\(\cos \Theta = \frac{Base}{Hypotenuse}\) …. (2)

By comparing (1) and (2)

We get,

Base=12 and

Hypotenuse = 15

1

Therefore,

By Pythagoras theorem,

AC2 = AB2+ BC2

Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

152 = 122 + BC2

BC2 = 152 – 122

BC2 = 225 – 144

BC 2= 81

BC = \(\sqrt{81}\)

BC = 9

Hence, Perpendicular side = 9

Now, \(\sin \Theta = \frac{Perpendicular}{Hypotenuse}\)

Therefore,

\(\sin \Theta = \frac{9}{15}\)

Now, cosec\(\Theta = \frac{1}{\sin \Theta }\)

Therefore,

cosec \(\Theta = \frac{Hypotenuse}{Perpendicular}\)

cosec\(\Theta = \frac{15}{9}\)

Now, \(\sec \Theta = \frac{1}{\cos \Theta }\)

Therefore,

\(\sec \Theta = \frac{Hypotenuse}{Base}\)

\(\sec \Theta = \frac{15}{12}\)

Now, \(\tan \Theta = \frac{Perpendicular}{Base}\)

Therefore,

\(\tan \Theta = \frac{9}{12}\)

Now, \(\cot \Theta = \frac{1}{\tan \Theta }\)

Therefore,

\(\cot \Theta = \frac{Base}{ Perpendicular}\)

\(\cot \Theta = \frac{12}{ 9}\)

 

2.)   In a \(\Delta\)ABC, right angled at B , AB – 24 cm , BC= 7 cm , Determine

(i) sin A , cos A

(ii) sin C, cos C

Sol.

(i) The given triangle is below:

 3

Given:   In \(\Delta\) ABC , AB= 24 cm

BC = 7cm

\(\angle ABC\) = 90o

To find: sin A, cos A

In this problem, Hypotenuse side is unknown

Hence we first find hypotenuse side by Pythagoras theorem

By Pythagoras theorem,

We get,

AC2 = AB2 + BC2

AC2 = 242 + 72

AC2 = 576 + 49

AC2= 625

AC = \(\sqrt{625}\)

AC= 25

Hypotenuse = 25

By definition,

\(\sin A = \frac{Perpendicular side opposite to \angle A}{Hypotenuse}\)

\(\sin A = \frac{BC}{AC}\)

\(\sin A = \frac{7}{25}\)

By definition,

\(\cos A = \frac{Base side adjacent to \angle A}{Hypotenuse}\)

\(\cos A = \frac{AB}{AC}\)

\(\cos A = \frac{24}{25}\)

Answer:

\(\sin A = \frac{7}{25}\), \(\cos A = \frac{24}{25}\)

 

(ii)  The given triangle is below:

3

Given:   In \(\Delta\) ABC , AB= 24 cm

BC = 7cm

\(\angle ABC\) = 90o

To find: sin C, cos C

In this problem, Hypotenuse side is unknown

Hence we first find hypotenuse side by Pythagoras theorem

By Pythagoras theorem,

We get,

AC2 = AB2 + BC2

AC2 = 242 + 72

AC2 = 576 + 49

AC2= 625

AC = \(\sqrt{625}\)

AC= 25

Hypotenuse = 25

By definition,

\(\sin C = \frac{Perpendicular side opposite to \angle C}{Hypotenuse}\)

\(\sin C = \frac{AB}{AC}\)

\(\sin C = \frac{24}{25}\)

By definition,

\(\cos C = \frac{Base side adjacent to \angle C}{Hypotenuse}\)

\(\cos A = \frac{BC}{AC}\)

\(\cos A = \frac{7}{25}\)

Answer:

\(\sin A = \frac{24}{25}\), \(\cos A = \frac{7}{25}\)

 

 3.) In the below figure, find tan P and cot R. Is tan P = cot R? 

  4                   

               

               To find, tan P, cot R

Sol.

   In the given right angled \(\Delta PQR\), length of side OR is unknown

Therefore , by applying Pythagoras theorem in \(\Delta PQR\)

We get,

PR2 = PQ2 + QR2

Substituting the length of given side PR and PQ in the above equation

132= 122 + QR2

QR2 = 132 – 122

QR2 = 169 – 144

QR2= 25

QR = \(\sqrt{25}\)

By definiton, we know that ,

\(\tan P = \frac{Perpendicular side opposite to \angle P }{Base side adjacent to \angle P}\)

\(\tan P = \frac{ QR }{ PQ }\)

\(\tan P = \frac{ 5 }{ 12 }\) …. (1)

Also, by definition, we know that

\(\cot R= \frac{Base\, side\, adjacent\, to\, \angle R}{Perpendicular\, \: side\, opposite\, to\, \angle R}\)

 

\(\cot R=\frac{QR}{PQ}\)

\(\cot R=\frac{5}{12}\) …. (2)

 

Comparing equation (1) ad (2), we come to know that that R.H.S of both the equation are equal.

Therefore, L.H.S of both equations is also equal

tan P = cot R

Answer:

Yes , tan P =cot R = \(\frac{5}{12}\)

 

4.)  If sin A = \(\frac{9}{41}\), Compute cos A and tan A.

Sol.

               Given:  \(\sin A= \frac{9}{41}\) …. (1)

To find: cos A, tan A

By definition,

\(\sin A= \frac{Perpendicular\, side\, opposite\, to\, \angle A}{Hypotenuse}\) …. (2)

By comparing (1) and (2)

We get ,

Perpendicular side = 9 and

Hypotenuse = 41

Now using the perpendicular side and hypotenuse we can construct \(\Delta ABC\) as shown below

5

Length of side AB is unknown is right angled \(\Delta ABC\) ,

To find the length of side AB, we use Pythagoras theorem,

Therefore, by applying Pythagoras theorem in \(\Delta ABC\) ,

We get,

AC2 = AB2 + BC2

412 = AB2 + 92

AB2 = 412 – 92

AB2 = 168 – 81

AB= 1600

AB = \(\sqrt{1600}\)

AB = 40

Hence, length of side AB= 40

Now

By definition,

\(\cos A =\frac{Base\, side\, adjacent\, to\, \angle A}{Hypotenuse}\)

\(\cos A =\frac{AB}{AC}\)

\(\cos A =\frac{40}{41}\)

Now,

By definition,

\(\tan A = \frac{Perpendicular\, side\, opposite\, to\, \angle A}{Base\, side\, adjacent\, to\, \angle A}\)

\(\tan A = \frac{BC}{AB}\)

\(\tan A = \frac{9}{40}\)

Answer:

\(\cos A =\frac{40}{41}\) , \(\tan A = \frac{9}{40}\)

 

5.)  Given 15cot A=8, find sin A and sec A.

Answer:

                 Given: 15cot A = 8

To find: sin A, sec A

Since 15 cot A =8

By taking 15 on R.H.S

We get,

\(\cot A = \frac{8}{15}\)

                By definition,

\(\cot A = \frac{1}{\tan A}\)

Hence,

\(\cot A = \frac{1}{\frac{Perpendicular\, side\, opposite\, to\, \angle A}{Base\, side\, adjacent\, to\, \angle A}}\)

\(\cot A= \frac{Base\, side\, adjacent\, to\, \angle A}{Perpendicular\, side\, opposite\, to\,  \angle A}\) …. (2)

Comparing equation (1) and (2)

We get,

Base side adjacent to \(\angle A\) = 8

Perpendicular side opposite to \(\angle A\) = 15

\(\Delta ABC\) can be drawn below using above information

6

Hypotenuse side is unknown.

Therefore, we find side AC of \(\Delta ABC\) by Pythagoras theorem.

So, by applying Pythagoras theorem to \(\Delta ABC\)

We get,

AC2 = AB2 +BC2

Substituting values of sides from the above figure

AC2 = 82 + 152

AC2 = 64 + 225

AC2 = 289

AC = \(\sqrt{289}\)

AC = 17

Therefore, hypotenuse =17

Now by definition,

\(\sin A=\frac{Perpendicular\, side\, opposite\, to\, \angle A}{Hypotenuse}\)

Therefore, \(\sin A=\frac{BC}{AC}\)

Substituting values of sides from the above figure

\(\sin A=\frac{ 15 }{ 17 }\)

By definition,

\(\sec A= \frac{1}{\cos A}\)

Hence,

\(\sec A=\frac{1}{\frac{Base\, side\, adjacent\, to\, \angle A}{Hypotenuse}}\)

\(\sec A= \frac{Hypotenuse}{Base\, side\, adjacent\, to\, \angle A}\)

Substituting values of sides from the above figure

\(\sec A= \frac{17}{8}\)

Answer:

 \(\sin A=\frac{ 15 }{ 17 }\), \(\sec A= \frac{17}{8}\)

 

  6.)        In \(\Delta PQR\) , right angled at Q, PQ = 4cm and RQ= 3 cm .Find the value of sin P,   sin R , sec P and sec R.

Sol.

                 Given:

\(\Delta PQR\) is right angled at vertex Q.

PQ= 4cm

RQ= 3cm

To find,

sin P, sin R , sec P , sec R  

                Given \(\Delta PQR\) is as shown below

7

Hypotenuse side PR is unknown.

Therefore, we find side PR of \(\Delta PQR\) by Pythagoras theorem

By applying Pythagoras theorem to \(\Delta PQR\)

We get,

PR2 = PQ2 +RQ2

Substituting values of sides from the above figure

PR2 = 42 +32

                        PR2 = 16 + 9

PR2 = 25

PR = \(\sqrt{25}\)

PR = 5

Hence, Hypotenuse =5

Now by definition,

\(\sin P =\frac{Perpendicular side opposite to \angle P}{Hypotenuse}\)

\(\sin P =\frac{RQ}{PR}\)

Substituting values of sides from the above figure

\(\sin P =\frac{3}{5}\)

Now by definition,

\(\sin R = \frac{Perpendicular\, side\, opposite\, to\, \angle R}{Hypotenuse}\)

\(\sin R = \frac{PQ}{PR}\)

Substituting the values of sides from above figure

\(\sin R = \frac{4}{5}\)

By definition,

\(\sec P=\frac{1}{\cos P}\)

\(\sec P=\frac{1}{\frac{Base\, side\, adjacent\, to\, \angle p}{Hypotenuse}}\)

\(\sec P=\frac{Hypotenuse}{Base\, side\, adjacent\, to\, \angle P}\)

Substituting values of sides from the above figure

\(\sec P =\frac{PR}{PQ}\)

\(\sec P =\frac{5}{4}\)

By definition,

\(\sec R =\frac{1}{\cos R}\)

\(\sec R =\frac{1}{\frac{Base\, side\, adjacent\, to\, \angle R}{Hypotenuse}}\)

\(\sec R=\frac{Hypotenuse}{Base\, side\, adjacent\, to\, \angle R}\)

Substituting values of sides from the above figure

\(\sec R = \frac{PR}{RQ}\)

\(\sec R = \frac{5}{3}\)

Answer:

\(\sin P =\frac{3}{5}\) , \(\sin R = \frac{4}{5}\),

                \(\sec P =\frac{5}{4}\), \(\sec R = \frac{5}{3}\)

 

7.)      If \(\cot \Theta = \frac{7}{8}\), evaluate

       (i)   \(\frac{1 + \sin \Theta\times 1 – \sin \Theta }{1 + \cos \Theta \times 1 – \cos \Theta  }\)

       (ii)  \(\cot^{2}\Theta\)

Sol.

                 Given: \(\cot \Theta = \frac{7}{8}\)

To evaluate:  \(\frac{1 + \sin \Theta\times 1 – \sin \Theta }{1 + \cos \Theta \times 1 – \cos \Theta  }\)

\(\frac{1 + \sin \Theta\times 1 – \sin \Theta }{1 + \cos \Theta \times 1 – \cos \Theta  }\) …( 1)

We know the following formula

(a + b)(a – b) = a2 – b2

By applying the above formula in the numerator of equation (1)

We get,

\((1 + sin \theta ) \times (1 – sin \theta ) = 1 – sin ^{2} \theta . . . . . (2)\;\;\;\;\;\;\; (Where, \;a = 1 \;and\; b = sin \theta )\)

Similarly,

By applying formula (a +b) (a – b) = a2 – b2 in the denominator  of equation (1).

We get,

\((1 + \cos \Theta )(1 – \cos \Theta )= 1^{2} – \cos ^{2}\Theta\) … (Where a=1 and b=   \(\cos \Theta\)

\((1 + \cos \Theta )(1 – \cos \Theta )= 1 – \cos ^{2}\Theta\) … (Where a=1 and b=   \(\cos \Theta\)

Substituting the value of numerator and denominator of equation (1) from equation (2), equation (3).

Therefore,

\(\frac{(1+ \sin \Theta ) (1 – \sin \Theta )}{(1 + \cos \Theta )(1 – \cos \Theta )}\) = \(\frac{1 – \sin ^{2}\Theta }{ 1 – \cos ^{2}\Theta }\) ….(4)

Since,

\(\cos ^{2}\Theta + \sin ^{2}\Theta = 1\)

Therefore,

\(\cos ^{2}\Theta = 1 – \sin ^{2\ }\Theta\)

Also, \(\sin ^{2}\Theta = 1 – \cos ^{2 }\Theta\)

Putting the value of \(1 – \sin ^{2}\Theta\) and \(1 – \cos ^{2}\Theta\) in equation (4)

We get,

\(\frac{(1 +\sin \Theta )( 1 – \sin \Theta )}{(1 +\cos \Theta )(1 – \cos \Theta )}\) = \(\frac{\cos ^{2}\Theta }{\sin ^{2}\Theta }\)

We know that, \(\frac{\cos \Theta }{\sin \Theta } = \cot \Theta\)

\(\frac{(1 +\sin \Theta )( 1 – \sin \Theta )}{(1 +\cos \Theta )(1 – \cos \Theta )}\) = \((\cot \Theta )^{2}\)

Since, it is given that \(\cot \Theta = \frac{7}{8}\)

Therefore,

\(\frac{(1 +\sin \Theta )( 1 – \sin \Theta )}{(1 +\cos \Theta )(1 – \cos \Theta )}\) = \((\frac{7}{8})^{2}\)

\(\frac{(1 +\sin \Theta )( 1 – \sin \Theta )}{(1 +\cos \Theta )(1 – \cos \Theta )}\)= \(\frac{7^{2}}{8^{2}}\)

\(\frac{(1 +\sin \Theta )( 1 – \sin \Theta )}{(1 +\cos \Theta )(1 – \cos \Theta )}\) = \(\frac{49}{64}\)

(ii) Given: \(\cot \Theta = \frac{7}{8}\)

To evaluate: \(\cot ^{2}\Theta\)

\(\cot \Theta = \frac{7}{8}\)

Squaring on both sides,

We get,

\((\cot \Theta )^{2}= (\frac{7}{8})^{2}\)

\((\cot \Theta )^{2}\)=\(\frac{49}{64}\)

Answer:

\(\frac{49}{64}\)

 

8.)    If \(3\cot A = 4\) , check whether  \(\frac{1 – \tan ^{2} A}{1 + \tan ^{2 }A}= \cos ^{2  }A – \sin ^{2}A\) or not.

Sol.

                 Given: 3cot A =4

To check whether \(\frac{1 – \tan ^{2} A}{1 + \tan ^{2 }A}= \cos ^{2  }A – \sin ^{2}A\) or not.

3cot A =4

Dividing by 3 on both sides,

We get,

cot A = \(\frac{4}{3}\) …. (1)

By definition,

\(\cot A=\frac{1}{\tan A}\)

Therefore,

\(\cot A = \frac{1}{\frac{Perpendicular\, side\, opposite\, to\, \angle A}{Base\, side\, adjacent\, to\, \angle A}}\)

\(\cot A= \frac{Base\, side\, adjacent\, to\, \angle A}{Perpendicular\, side\, opposite\, to\, \angle A}\) …. (2)

Comparing (1) and (2)

We get,

Base side adjacent to \(\angle A\) = 4

Perpendicular side opposite to \(\angle A\) = 3

Hence \(\Delta ABC\) is as shown in figure below

8

In \(\Delta ABC\) , Hypotenuse is unknown

Hence, it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem in\(\Delta ABC\)

We get

AC2= AB2 +BC2

Substituting the values of sides from the above figure

AC2 =42 + 32

AC2 = 16 +9

AC2 = 25

AC = \(\sqrt{25}\)

AC = 5

Hence, hypotenuse= 5

To check whether

To check whether \(\frac{1 – \tan ^{2} A}{1 + \tan ^{2 }A}= \cos ^{2  }A – \sin ^{2}A\) or not.

We get thee values of tan A , cos A , sin A

By definition,

tan A = \(\frac{1}{\cot A}\)

Substituting the value of cot A from equation (1)

We get,

tan A  = \(\frac{1}{4}\)

tan A = \(\frac{3}{4}\) …. (3)

Now by definition,

\(\cos A = \frac{Base\, side\, adjacent\, to\, \angle A}{Hypotenuse}\)

\(\cos A = \frac{AB}{AC}\)

Substituting the values of sides from the above figure

\(\cos A= \frac{4}{5}\) …. (5)

Now we first take L.H.S of equation \(\frac{1 – \tan ^{2} A}{1 + \tan ^{2 }A}= \cos ^{2  }A – \sin ^{2}A\)

L.H.S = \(\frac{1 – \tan ^{2}A}{1 + \tan ^{2}A}\)

Substituting value of tan A from equation (3)

We get,

L.H.S= \(\frac{1 – (\frac{3}{4})^{2}}{1 + (\frac{3}{4})^{2}}\)

\(\frac{1 – \tan ^{2}A}{1 + \tan ^{2}A}\)=\(\frac{1 – (\frac{3}{4})^{2}}{1 + (\frac{3}{4})^{2}}\)

\(\frac{1 – \tan ^{2}A}{1 + \tan ^{2}A}\) = \(\frac{1 – \frac{9}{16}}{ 1+ \frac{9}{16}}\)

Taking L.C.M on both numerator and denominator

We get,

\(\frac{1 – \tan ^{2}A}{1 + \tan ^{2}A}\) = \(\frac{\frac{16 – 9}{16}}{\frac{16 + 9}{16}}\)

\(\frac{1 – \tan ^{2}A}{1 + \tan ^{2}A}\)= \(\frac{7}{25}\) …. (6)

Now we take R.H.S of equation whether \(\frac{1 – \tan ^{2} A}{1 + \tan ^{2 }A}= \cos ^{2  }A – \sin ^{2}A\)

R.H.S = \(\cos ^{2}A – \sin ^{2}A\)

Substituting value of sin A and cos A from equation (4) and (5)

We get,

R.H.S= \((\frac{4}{5})^{2} – {(\frac{3}{5}^{2})}\)

\(\cos ^{2}A – \sin ^{2}A\)= \((\frac{4}{5})^{2} – {(\frac{3}{5}^{2})}\)

\(\cos ^{2}A – \sin ^{2}A\) = \(\frac{16}{25} – \frac{9}{25}\)

\(\cos ^{2}A – \sin ^{2}A\) = \(\frac{16 – 9}{25}\)

\(\cos ^{2}A – \sin ^{2}A\) = \(\frac{7}{25}\) ….(7)

Comparing (6) and (7)

We get.

\(\frac{1 – \tan ^{2} A}{1 + \tan ^{2 }A}= \cos ^{2  }A – \sin ^{2}A\)

Answer:

Yes, \(\frac{1 – \tan ^{2} A}{1 + \tan ^{2 }A}= \cos ^{2  }A – \sin ^{2}A\)

 

 9.)  If \(\tan \Theta =\frac{a}{b}\) , find the value of \(\frac{\cos \Theta + \sin \Theta }{\cos \Theta – \sin \Theta }\).

Sol.

Given:

\(\tan \Theta = \frac{a}{b}\) …. (1)

Now, we know that \(\tan \Theta = \frac{\sin \Theta }{\cos \Theta }\)

Therefore equation (1) become as follows

\(\frac{\sin \Theta }{\cos \Theta }\) =\(\frac{a}{b}\)

Now, by applying invertendo

We get,

\(\frac{\cos \Theta }{\sin \Theta } =\frac{ b }{ a }\)

Now by applying Componendo – dividendo

We get,

\(\frac{\cos \Theta +\sin \Theta }{\cos \Theta – \sin \Theta } = \frac{b+ a }{b – a}\)

Therefore,

\(\frac{\cos \Theta +\sin \Theta }{\cos \Theta – \sin \Theta } = \frac{b+ a }{b – a}\)

 

10.)    If \(3\tan \Theta =4\) , find the value of \(\frac{4\cos \Theta – \sin \Theta }{2\cos \Theta + \sin \Theta }\) 

Sol.

Given: If \(3\tan \Theta =4\)

Therefore,

\(\tan \Theta = \frac{4}{3}\) …. (1)

Now, we know that \(\tan \Theta = \frac{\sin \Theta }{\cos \Theta }\)

Therefore equation (1) becomes

\(\frac{\sin \Theta }{\cos \Theta }=\frac{4}{3}\) ….(2)

Now, by applying Invertendo to equation (2)

We get,

\(\frac{\cos \Theta }{\sin \Theta }=\frac{3}{4}\) …. (3)

Now, multiplying by 4 on both sides

We get

\(4\times \frac{\cos \Theta }{\sin \Theta }= 4\times \frac{3}{4}\)

Therefore

\(\frac{4\cos \Theta – \sin \Theta }{\sin \Theta } = \frac{3 – 1}{1}\)

\(\frac{4\cos \Theta – \sin \Theta }{\sin \Theta } = \frac{2}{1}\) …. (4)

Now, multiplying by 2 on both sides of equation (3)

We get,

\(\frac{2\cos \Theta }{\sin \Theta }=\frac{3}{2}\)

Now by applying componendo in above equation

\(\frac{2\cos \Theta + \sin \Theta }{\sin \Theta }=\frac{3 + 2}{2}\)

\(\frac{2\cos \Theta + \sin \Theta }{\sin \Theta }=\frac{5}{2}\) ….(5)

We get,

\(\frac{\frac{4 \cos \Theta – \sin \Theta }{\sin \Theta }}{\frac{2\cos \Theta +\sin \Theta }{\sin \Theta }} = \frac{\frac{2}{1}}{\frac{5}{2}}\)

Therefore,

\(\frac{4\cos \Theta – \sin \Theta }{\sin \Theta } \times \frac{\sin \Theta }{2 \cos \Theta + \sin \Theta }= \frac{2}{1}\times \frac{2}{5}\)

Therefore, on L.H.S \(\sin \Theta\) cancels and we get,

\(\frac{4\cos \Theta – \sin \Theta }{2 \cos \Theta + \sin \Theta } = \frac{2}{1}\times \frac{2}{5}\)

Therefore,

\(4\cos \Theta – \sin \Theta = 4\)

 

11.)   If \(3\cot \Theta = 2\), find the value of \(\frac{4 \sin \Theta – 3 \cos \Theta }{2           \sin \Theta + 6\cos \Theta }\)                                                                                                                                                                                       Sol.

Given:

\(3\cot \Theta = 2\)

Therefore,

\(\cot \Theta = \frac{2}{3}\) …. (1)

Now, we know that \(\cot \Theta = \frac{\cos \Theta }{\sin \Theta }\)

Therefore equation (1) becomes

\(\frac{\cos \Theta }{\sin \Theta }=\frac{2}{3}\) ….(2)

Now , by applying invertendo to equation (2)

\(\frac{\sin \Theta }{\cos \Theta }=\frac{3}{2}\) ….(3)

Now, multiplying by \(\frac{4}{3}\) on both sides,

We get,

\(\frac{4}{3}\times \frac{\sin \Theta }{\cos \Theta}= \frac{4}{3}\times \frac{3}{2}\)

Therefore, 3 cancels out on R.H.S and

We get,

\(\frac{4\sin \Theta }{3\cos \Theta }= \frac{2}{1}\)

Now by applying invertendo dividendo in above equation

We get,

\(\frac{4\sin \Theta – 3 \cos \Theta }{3\cos \Theta }= \frac{2 – 1}{1}\)

\(\frac{4\sin \Theta – 3 \cos \Theta }{3\cos \Theta }= \frac{1}{1}\) ….(4)

Now, multiplying by \(\frac{2}{6}\) on both sides of equation (3)

We get,

\(\frac{2}{6}\times \frac{\sin \Theta }{\cos \Theta }= \frac{2}{6}\times \frac{3}{2}\)

Therefore, 2 cancels out on R.H.S and

We get,

\(\frac{2\sin \Theta }{6\cos \Theta }= \frac{3}{6}\)

\(\frac{2\sin \Theta }{6\cos \Theta }= \frac{1}{2}\)

Now by applying componendo in above equation

We get,

\(\frac{2\cos \Theta + 6 \sin \Theta }{6\sin \Theta }= \frac{1 + 2}{2}\)

\(\frac{2\cos \Theta + 6 \sin \Theta }{6\sin \Theta }= \frac{3}{2}\) ….(5)

Now, by dividing equation (4) by (5)

We get,

\(\frac{\frac{4 \sin \Theta – 3 \cos \Theta }{3 \sin \Theta }}{\frac{2 \cos \Theta + 6\sin \Theta }{6 \sin \Theta }}= \frac{\frac{1}{1}}{\frac{3}{2}}\)

Therefore,

\(\frac{4 \sin \Theta – 3 \cos \Theta }{3 \sin \Theta }\times \frac{6 \sin \Theta }{2 \cos \Theta + 6\sin \Theta}= {\frac{1}{1}} \times {\frac{2}{3}}\)

\(\frac{4 \sin \Theta – 3 \cos \Theta }{3 \sin \Theta }\times \frac{2\times 3 \sin \Theta }{2 \cos \Theta + 6\sin \Theta}= {\frac{1}{1}} \times {\frac{2}{3}}\)

Therefore, on L.H.S (3 \(\sin \Theta\)) cancels out and we get,

\(\frac{2 \times 4 \sin \Theta – 3 \cos \Theta }{2 \cos \Theta + 6\sin \Theta}= {\frac{1}{1}} \times {\frac{2}{3}}\)

Now, by taking 2 in the numerator of L.H.S on the R.H.S

We get,

\(\frac{ 4 \sin \Theta – 3 \cos \Theta }{2 \cos \Theta + 6\sin \Theta}= \frac{2}{3\times 2}\)

Therefore, 2 cancels out on R.H.S and

We get,

\(\frac{ 4 \sin \Theta – 3 \cos \Theta }{2 \cos \Theta + 6\sin \Theta}= \frac{1}{3}\)

Hence answer,

\(\frac{ 4 \sin \Theta – 3 \cos \Theta }{2 \cos \Theta + 6\sin \Theta}= \frac{1}{3}\)

 

12.)    If \(\tan \Theta = \frac{a}{b}\), prove that \(\frac{a \sin \Theta – b \cos \Theta }{a \sin \Theta + b \cos \Theta }= \frac{ a ^{2} – b ^{2}}{a ^{2 } + b^{2}}\)

Sol.

Given:

\(\tan \Theta \frac{a}{b}\) …. (1)

Now, we know that \(\tan \Theta = \frac{\sin \Theta }{\cos \Theta }\)

Therefore equation (1) becomes

\(\frac{\sin \Theta }{\cos \Theta }=\frac{a}{b}\) ….(2)

Now, by multiplying by \(\frac{a}{b}\) on both sides of equation (2)

We get,

\(\frac{a}{b}\times \frac{\sin \Theta }{\cos \Theta } = \frac{a}{b}\ \times \frac{a}{b}\)

Therefore,

\(\frac{a\sin \Theta }{b\cos \Theta } = \frac{a^{2}}{b^{2}}\) ….(3)

Now by applying dividendo in above equation (3)

We get,

\(\frac{a\sin \Theta – b\cos \Theta }{b\cos \Theta } = \frac{a^{2} – b^{2}}{b^{2}}\) ….(4)

Now by applying componendo in equation (3)

We get,

\(\frac{a\sin \Theta + b\cos \Theta }{b\cos \Theta } = \frac{a^{2} + b^{2}}{b^{2}}\) ….(5)

Now, by dividing equation (4) by equation (5)

We get,

\(\frac{\frac{a\sin \Theta – b \cos \Theta }{b \cos \Theta }}{\frac{a \sin \Theta + b \cos \Theta }{b \cos \Theta }}= \frac{\frac{a^{2}- b^{2}}{b^{2}}}{\frac{a^{2}+ b^{2}}{b^{2}}}\)

Therefore,

\(\frac{a\sin \Theta – b\cos \Theta }{b \cos \Theta }\times \frac{b \cos \Theta }{a \sin \Theta + b\cos \Theta }= \frac{a^{2}- b^{2}}{b^{2}}\times \frac{b^{2}}{a^{2}+b^{2}}\)

Therefore, \(b \cos \Theta\) and b2 cancels on L.H.S and R.H.S respectively

\(\frac{a\sin \Theta – b\cos \Theta }{a \sin \Theta + b\cos \Theta }= \frac{a^{2}- b^{2}}{a^{2}+b^{2}}\)

Hence, it is proved that

 \(\frac{a\sin \Theta – b\cos \Theta }{a \sin \Theta + b\cos \Theta }= \frac{a^{2}- b^{2}}{a^{2}+b^{2}}\)

13.)   If \(\sec \Theta = \frac{13}{5}\), show that \(\frac{2 \sin \Theta – 3 \cos \Theta }{4 \sin \Theta – 9\cos \Theta }= 3\)

 Sol.

Given:

\(\sec \Theta = \frac{13}{5}\)

To show that \(\frac{2 \sin \Theta – 3 \cos \Theta }{4 \sin \Theta – 9\cos \Theta }= 3\)

Now, we know that \(\cos \Theta = \frac{1}{\sec \Theta }\)

Therefore,

\(\cos \Theta = \frac{1}{\frac{13}{5}}\)

Therefore,

\(\cos \Theta = \frac{5}{13}\) …. (1)

Now, we know that

\(\cos \Theta = \frac{Base\, side\, adjacent\, to \, \angle \Theta }{Hypotenuse}\)

Now, by comparing equation (1) and(2)

We get,

Base side adjacent to \(\angle \Theta\) = 5

And

Hypotenuse =13

9

Therefore from above figure

Base side BC = 5

Hypotenuse AC = 13

Side AB is unknown. It can be determined by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

AC2 = AB2 + BC2

Therefore by substituting the values of known sides

We get,

132 = AB2 + 52

Therefore,

AB2= 132 – 52

AB2= 169 – 25

AB2 = 144

AB = \(\sqrt{144}\)

Therefore,

AB = 12 …. (3)

Now, we know that

\(\sin \Theta = \frac{AB}{AC}\)

\(\sin \Theta = \frac{12}{13}\) …. (4)

Now L.H.S of the equation to be proved is as follows

L.H.S = \(\frac{2 \sin \Theta – 3 \tan \Theta }{4 \sin \Theta – 3 \cos \Theta }\)

Substituting the value \(\cos \Theta\) of \(\sin \Theta\)and from equation (1) and (4) respectively

We get,

\(\frac{2 \times \frac{12}{13} – 3 \times \frac{5}{13} }{4 \times \frac{12}{13} – 9 \times \frac{5}{13}}\)

Therefore,

L.H.S = \(\frac{2 \times 12 – 3 \times 5}{4 \times 12 – 9 \times 5}\)

L.H.S = \(\frac{24 – 15}{48 – 45}\)

L.H.S= \(\frac{9}{3}\)

L.H.S= 3

Hence proved that,

  \(\frac{2 \sin \Theta – 3 \tan \Theta }{4 \sin \Theta – 3 \cos \Theta }\)= 3

 

14.)   If  \(\cos \Theta = \frac{12}{13}\) , show that \(\sin \Theta (1 – \tan \Theta )= \frac{35}{156}\)

Sol.

                Given: \(\cos \Theta = \frac{12}{13}\) …. (1)

To show that \(\sin \Theta (1 – \tan \Theta )= \frac{35}{156}\)

Now we know that  \(\cos \Theta = \frac{Base\, side\, adjacent\, to\, \angle\, \Theta }{Hypotenuse}\) ….(2)

Therefore, by comparing equation (1) and (2)

We get,

Base side adjacent to \(\angle \Theta\) = 12

And

Hypotenuse = 13

10

Therefore from above figure

Base side BC= 12

Hypotenuse AC= 13

Side AB is unknown and it can be determined by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

AC2= AB2 + BC2

Therefore by substituting the values of known sides

We get,

132= AB2 + 122

Therefore,

AB2= 132– 122

AB2= 169 – 144

AB =25

AB= \(\sqrt{25}\)

Therefore,

AB = 5 …. (3)

Now, we know that

\(\sin \Theta = \frac{Perpendicular\, side\, opposite\, to\, \angle \Theta }{Hypotenuse}\)

Now from figure (a)

We get,

\(\sin \Theta = \frac{AB}{AC}\)

Therefore,

\(\sin \Theta = \frac{5}{12}\) …. (5)

Now L.H.S of the equation to be proved is as follows

L.H.S of the equation to be proved is as follows

L.H.S = \(\sin \Theta (1 – \tan \Theta]\) …. (6)

Substituting the value of \(\sin \Theta\) and \(\tan \Theta\) from equation (4) and (5)

We get,

L.H.S = \(\frac{5}{13}(1-\frac{5}{12})\)

Taking L.C.M inside the bracket

We get,

L.H.S= \(\frac{5}{13}(\frac{1\times 12}{1\times 12}-\frac{5}{12})\)

Therefore,

L.H.S= \(\frac{5}{13}(\frac{12 – 5 }{12})\)

L.H.S = \(\frac{5}{13}(\frac{7}{12})\)

Now by opening the bracket and simplifying

We get,

L.H.S = \(\frac{5\times 7}{13\times 12}\)

L.H.S= \(\frac{35}{136}\)

From equation (6) and (7) ,it can be shown that

that \(\sin \Theta (1 – \tan \Theta )\) = \(\frac{35}{136}\)

15.)    If \(\cot \Theta = \frac{1}{\sqrt{3}}\) , show that \(\frac{1 – \cos  ^{2}\Theta }{2- \sin ^{2}\Theta }= \frac{3}{5}\)

Sol.

                Given: \(\cot \Theta = \frac{1}{\sqrt{3}}\) …. (1)

To show that \(\frac{1 – \cos  ^{2}\Theta }{2- \sin ^{2}\Theta }= \frac{3}{5}\)

Now, we know that \(\cot \Theta = \frac{1}{\tan \Theta }\)

Since \(\tan \Theta = \frac{Perpendicular\, side \, opposite \, to\, \angle \Theta }{Base\, side \, adjacent\, to\, \angle \Theta }\)

Therefore,

\(\tan \Theta =\frac{1}{\frac{Perpendicular\, side \, opposite \, to\, \angle \Theta }{Base\, side \, adjacent\, to\, \angle \Theta }}\)

Therefore,

\(\cot \Theta =\frac{Base\, side \, adjacent\, to\, \angle \Theta }{Perpendicular\, side \, opposite \, to\, \angle \Theta }\) …. (2)

Comparing Equation (1) and (2)

We get.

Base side adjacent to \(\angle \Theta\) = 1

Perpendicular side opposite to \(\angle \Theta\) = \(\sqrt{3}\)

Therefore, triangle representing angle \(\sqrt{3}\) is as shown below

11

Therefore, by substituting the values of known sides

We get,

AC2 = \((\sqrt{3})^{2}\) + 12

Therefore,

AC2 = 3 + 1

AC2= 4

AC= \(\sqrt{4}\)

Therefore,

AC = 2 …. (3)

Now, we know that

\(\sin \Theta = \frac{Perpendicular\, side\, opposite\, to\, \angle \Theta }{Hypotenuse}\)

Now from figure (a)

\(\sin \Theta = \frac{AB}{AC}\)

Therefore from figure (a) and equation (3),

\(\sin \Theta = \frac {\sqrt 3}{2}\)

Now we know that

\(\cos \Theta \frac{Base\, side \, adjacent\, to \, \angle \Theta }{Hypotenuse}\)

Now from figure (a)

We get,

\(\frac{BC}{AC}\)

Therefore from figure (a) and equation (3),

\(\cos \Theta = \frac{1}{2}\) …. (5)

Now, L.H.S of the equation to be proved is as follows

L.H.S = \(\frac{1 – \cos ^{2}\Theta }{2 – \sin ^{2}\Theta }\)

Substituting the value of from equation (4) and (5)

We get,

L.H.S = \(\frac{1 – (\frac{1}{2})^{2}}{2 – (\frac{\sqrt{3}}{2})^{2}}\)

L.H.S = \(\frac{1 – \frac{1}{4}}{2-\frac{3}{4}}\)

Now by taking L.C.M in numerator as well as denominator

We get,

L.H.S= \(\frac{\frac{(4\times 1) – 1}{4}}{\frac{(4\times 2) – 3}{4}}\)

Therefore,

L.H.S = \(\frac{\frac{4 – 1}{4}}{\frac{8 – 3}{4}}\)

L.H.S = \(\frac{3}{4}\times \frac{4}{5}\)

L.H.S = \(\frac{3}{5}\) = R.H.S

Therefore,

\(\frac{1 – \cos  ^{2}\Theta }{2- \sin ^{2}\Theta }= \frac{3}{5}\)

 

16.)   If \(\tan \Theta = \frac{1}{\sqrt{7}}\), then show that \(\frac{cosec^{2}\Theta – \sec ^{2}\Theta }{cosec^{2}\Theta + \sec ^{2}\Theta }\) = \(\frac{3}{4}\)

Sol.

                Given: \(\tan \Theta = \frac{1}{\sqrt{7}}\) …. (1)

To show that          \(\frac{cosec^{2}\Theta – \sec ^{2}\Theta }{cosec^{2}\Theta + \sec ^{2}\Theta }\) = \(\frac{3}{4}\)

Now, we know that

Since, \(\tan \Theta = \frac{Perpendicular\, side\, oposite\, to\, \angle \Theta }{Base\, side\, adjacent\, to\, \angle \Theta }\) ….(2)

Therefore,

Comparing equation (1) and (2)

We get.

Perpendicular side opposite to \(\angle \Theta\) = 1

Base side adjacent to \(\angle \Theta\) = \(\sqrt{7}\)

Therefore, Triangle representing \(\angle \Theta\) is shown below

12

Hypotenuse AC is unknown and it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

AC2= AB2 + BC2

Therefore by substituting the values of known sides

We get,

AC2= (1)2 + \((\sqrt{7})^{2}\)

Therefore,

AC 2 = 1 +7

AC2 = 8

AC = \(\sqrt{8}\)

AC = \(\sqrt{2\times 2\times 2}\)

Therefore,

AC = \(2\sqrt{2}\) …. (3)

Now we know that

\(\sin \Theta = \frac{Perpendicular \, side\, opposite\, to\, \angle \Theta }{Hypotenuse}\)

\(\sin \Theta = \frac{AB }{AC}\)

\(\sin \Theta = \frac{1 }{2\sqrt{2}}\) …. (4)

Now, we know that cosec \(\Theta = \frac{1}{\sin \Theta }\)

Therefore, from equation (4)

We get,

cosec \(\Theta = 2\sqrt{2}\) …. (5)

Now, we know that

\(\cos \Theta = \frac{Base\, side \, adjacent\, to\, \angle \Theta }{Hypotenuse}\)

Now from figure (a)

We get,

\(\cos \Theta = \frac{BC}{AC}\)

Therefore from figure (a) and equation (3)

\(\cos \Theta = \frac{\sqrt{7}}{2\sqrt{2}}\) …. (6)

Now we know that \(\sec \Theta = \frac{1 }{\cos \Theta }\)

Therefore, from equation (6)

We get,

\(\sec \Theta = \frac{1}{\frac{\sqrt{7}}{2\sqrt{2}}}\)

\(\sec \Theta= \frac{2\sqrt{2}}{\sqrt{7}}\) …. (7)

Now, L.H.S of the equation to be proved is as follows

L.H.S = \(\frac{cosec^{2}\Theta – \sec ^{2}\Theta }{cosec^{2}\Theta + \sec ^{2}\Theta }\)

Substituting the value of cosec\(\Theta\) and\(\sec \Theta\) from equation (6) and (7)

We get,

L.H.S = \(\frac{\left [\left (2\sqrt{2} \right ) \right ]^{2} – \left (\frac{2\sqrt{2}}{\sqrt{7}} \right )^{2}}{\left [\left (2\sqrt{2} \right ) \right ]^{2} + \left (\frac{2\sqrt{2}}{\sqrt{7}} \right )^{2}}\)

L.H.S= \(\frac{\left (8 \right )- \left (\frac{8}{7} \right )}{\left (8 \right ) + \left (\frac{8}{7} \right )}\)

Therefore,

\(\frac{\frac{56 – 8}{7}}{\frac{56 + 8 }{7}}\)

L.H.S = \(\frac{\frac{48}{7}}{\frac{64 }{7}}\)

Therefore,

L.H.S = \(\frac{48}{64}\)

L.H.S = \(\frac{3}{4}\) = R.H.S

Therefore,

\(\frac{cosec^{2}\Theta – \sec ^{2}\Theta }{cosec^{2}\Theta + \sec ^{2}\Theta }\) = \(\frac{3}{4}\)

Hence proved that

\(\frac{cosec^{2}\Theta – \sec ^{2}\Theta }{cosec^{2}\Theta + \sec ^{2}\Theta }\) = \(\frac{3}{4}\)

 

17.)    If \(\sec \Theta = \frac{5}{4}\),find the value of \(\frac{\sin \Theta – 2\cos \Theta }{\tan \Theta – \cot \Theta }\)

Sol.

                Given: \(\sec \Theta = \frac{5}{4}\) …. (1)

To find the value of \(\frac{\sin \Theta – 2\cos \Theta }{\tan \Theta – \cot \Theta }\)

Now we know that \(\sec \Theta = \frac{1}{\cos \Theta }\)

Therefore,

\(\cos \Theta = \frac{1}{\sec \Theta }\)

Therefore from equation (1)

\(\cos \Theta = \frac{1}{5}\)

\(\cos \Theta = \frac{4}{5}\) …. (2)

Also, we know that \(\cos ^{2}\Theta + \sin ^{2}\Theta = 1\)

Therefore,

\(\sin ^{2}\Theta = 1 – \cos ^{2}\Theta\)

\(\sin \Theta =\sqrt{ 1 – \cos ^{2}\Theta }\)

Substituting the value of \(\cos \Theta\)from equation (2)

We get,

\(\sin \Theta = \sqrt{1 -\left ( \frac{4}{5} \right )^{2}}\)

= \(\sqrt{1-\frac{16}{25}}\)

= \(\frac{9}{25}\)

= \(\frac{3}{5}\)

Therefore,

\(\sin \Theta = \frac{3}{5}\) …. (3)

Also, we know that

\(\sec ^{2}\Theta = 1 + \tan ^{2}\Theta\)

Therefore,

\(\tan ^{2}\Theta = \left (\frac{5}{4} \right )^{2}- 1\)

\(\tan \Theta =\left ( \sqrt{}\frac{9}{16} \right )\)

Therefore,

\(\tan \Theta =\frac{3}{4}\) …. (4)

Also, \(\cot \Theta= \frac{1}{\tan \Theta }\)

Therefore from equation (4)

We get,

\(\cot \Theta= \frac{4}{3 }\) …. (5)

Substituting the value of \(\cos \Theta\), \(\cot \Theta\) and \(\tan \Theta\) from the equation (2),(3),(4) and (5) respectively in the expression below

\(\frac{\sin \Theta – 2\cos \Theta }{\tan \Theta – \cot \Theta }\)

We get,

\(\frac{\sin \Theta – 2\cos \Theta }{\tan \Theta – \cot \Theta }\)= \(\frac{\frac{3}{5}- 2\left (\frac{4}{5} \right )}{\frac{3}{4} – \frac{4}{3}}\)

=\(\frac{12}{7}\)

Therefore, \(\frac{\sin \Theta – 2\cos \Theta }{\tan \Theta – \cot \Theta }\)= \(\frac{12}{7}\)

 

18.)   If \(\sin \Theta = \frac{12}{13}\) , find the value of \(\frac{2\sin \Theta \cos \Theta }{\cos ^{2}\Theta – \sin ^{2}\Theta }\)

Sol.

                Given: \(\sin \Theta = \frac{12}{13}\) …. (1)

To, find the value of \(\frac{2\sin \Theta \cos \Theta }{\cos ^{2}\Theta – \sin ^{2}\Theta }\)

Now, we know the following trigonometric identity

cosec2 \(\Theta = 1 + \tan ^{2}\Theta\)

Therefore, by substituting the value of \(\tan \Theta\) from equation (1)

We get,

cosec2 \(\Theta = 1 +\left (\frac{12}{13} \right )^{2}\)

= \(1 + \frac{12^{2}}{13^{2}}\)

= \(1 + \frac{144}{169}\)

By taking L.C.M on the R.H.S

We get,

cosec2\(\Theta = \frac{169 + 144 }{169}\)

= \(\frac{313}{169}\)

Therefore

cosec\(\Theta=\sqrt{ \frac{313}{169}}\)

=\(\Theta =\frac{\sqrt{313}}{13}\)

Therefore

cosec\(\Theta\) = \(\Theta =\frac{\sqrt{313}}{13}\) …. (2)

Now, we know that

\(cosec\Theta\) = \(\frac{1}{\sin \Theta }\)

\(\sin \Theta = \frac{1}{\frac{\sqrt{313}}{13}}\)

Therefore

\(\sin \Theta = \frac{13}{\sqrt{313}}\) …. (3)

Now, we know the following trigonometric identity

\(\cos ^{2}\Theta + \sin ^{2}\Theta = 1\)

Therefore,

\(\cos ^{2}\Theta = 1 – \sin ^{2}\Theta\)

Now by substituting the value of \(\sin \Theta\) from equation (3)

We get,

\(\cos ^{2}\Theta = 1 – \left (\frac{13}{\sqrt{313}} \right )^{2}\)

=\(1 – \frac{169}{313}\)

Therefore, by taking L.C.M on R.H.S

We get,

\(\cos ^{2}\Theta = \frac{144}{313}\)

Now, by taking square root on both sides

We get,

\(\cos \Theta = \frac{12}{\sqrt{313}}\)

Therefore,

\(\cos \Theta = \frac{12}{\sqrt{313}}\) …. (4)

Substituting the value of \(\sin \Theta\) and \(\cos \Theta\) from equation (3) and (4) respectively in the equation below

\(\frac{2\sin \Theta \cos \Theta }{\cos ^{2}\Theta – \sin ^{2}\Theta }\)

Therefore,

\(\frac{2\sin \Theta \cos \Theta }{\cos ^{2}\Theta – \sin ^{2}\Theta }\)= \(\frac{2 \times \frac{13 }{\sqrt{313}}\times \frac{12}{\sqrt{313}}}{(\frac{13}{\sqrt{313}})^{2} – (\frac{12}{\sqrt{313}})^{2}}\)

=\(\frac{\frac{312}{313}}{\frac{25}{313}}\)

\(\frac{312}{25}\)

Therefore

 \(\frac{2\sin \Theta \cos \Theta }{\cos ^{2}\Theta – \sin ^{2}\Theta }\)=

\(\frac{312}{25}\)

19.)    If \(\cos \Theta = \frac{3}{5}\) , find the value of \(\frac{\sin \Theta – \frac{1}{\tan \Theta }}{2\tan \Theta }\)   

Sol.

Given: \(\cos \Theta = \frac{3}{5}\) …. (1)

To find the value of \(\frac{\sin \Theta – \frac{1}{\tan \Theta }}{2\tan \Theta }\)   

Now we know the following trigonometric identity

\(\cos ^{2}\Theta + \sin ^{2} \Theta = 1\)

Therefore by substituting the value of \(\cos \Theta\) from equation (1)

We get,

\((\frac{3}{5})^{2} + \sin ^{2}\Theta = 1\)

Therefore,

\(\sin ^{2}\Theta = 1 – (\frac{3}{5})^{2}\)

\(\sin ^{2}\Theta = 1 – (\frac{9}{25})\)

\(\sin ^{2}\Theta = \frac{25 – 9}{25}\)

\(\sin ^{2}\Theta = \frac{16}{25}\)

Therefore by taking square root on both sides

We get,

\(\sin \Theta = \frac{4}{5}\) …. (2)

Now, we know that

\(\tan \Theta = \frac{\sin \Theta }{\cos \Theta }\)

Therefore by substituting the value of \(\sin \Theta\) and \(\cos \Theta\) from equation (2) and (1) respectively

We get,

\(\tan \Theta = \frac{\frac{4}{5}}{\frac{3}{5}}= \frac{4}{3}\) …. (4)

Now, by substituting the value of \(\sin \Theta\) and of \(\tan\Theta\) from equation (2) and equation (4) respectively in the expression below

\(\frac{\sin \Theta – \frac{1}{\tan \Theta }}{2\tan \Theta }\)   

We get,

\(\frac{\sin \Theta – \frac{1}{\tan \Theta }}{2\tan \Theta }\)   =\(\frac{\frac{4}{5} – \frac{1}{4}}{2 \times \frac{4}{3}}\)

\(\frac{\sin \Theta – \frac{1}{\tan \Theta }}{2\tan \Theta }\)  = \(\frac{\frac{16}{20} – \frac{15}{20}}{\frac{8}{3}}\)

\(\frac{\sin \Theta – \frac{1}{\tan \Theta }}{2\tan \Theta }\)   = \(\frac{3}{160}\)

Therefore,

\(\frac{\sin \Theta – \frac{1}{\tan \Theta }}{2\tan \Theta }\) = \(\frac{3}{160}\)

20.)   If \(\sin \Theta = \frac{3}{5}\) ,  find the value of \(\frac{\cos \Theta – \frac{1}{\tan \Theta }}{2 \cot \Theta }\)

Sol.

              Given:

\(\sin \Theta = \frac{3}{5}\) …. (1)

To find the value of \(\frac{\cos \Theta – \frac{1}{\tan \Theta }}{2 \cot \Theta }\)

Now, we know the following trigonometric identity

\(\cos ^{2}\Theta + \sin ^{2} \Theta = 1\)

Therefore by substituting the value of \(\cos \Theta\) from equation (1)

We get,

\(\cos ^{2}\Theta + (\frac{3}{5})^{2} = 1\)

Therefore,

\(\cos ^{2}\Theta = 1 – (\frac{3}{5})^{2}\)

\(\cos ^{2}\Theta = 1 – \frac{9}{25}\)

Now by taking L.C.M

We get,

\(\cos ^{2}\Theta = \frac{25 – 9}{25}\)

\(\cos ^{2}\Theta = \frac{25 – 9}{25}\)

Therefore, by taking square roots on both sides

We get,

\(\cos \Theta = \frac{4}{5}\)

Therefore,

\(\cos \Theta = \frac{4}{5}\) …. (2)

Now we know that

\(\tan \Theta = \frac{\sin \Theta }{\cos \Theta }\)

Therefore by substituting the value of \(\sin \Theta\) and \(\cos\Theta\) from equation (1) and (2) respectively

We get,

\(\tan \Theta =\frac{ \frac{3}{5}}{\frac{4}{5}}\)

\(\tan \Theta =\frac{3}{4}\) …. (3)

Also, we know that

\(\cot \Theta = \frac{1}{\tan \Theta }\)

Therefore from equation (3)

We get,

\(\cot \Theta = \frac{1}{\frac{3}{4} }\)

\(\cot \Theta = {\frac{4}{3} }\) …. (4)

Now by substituting the value of \(\cos \Theta\), \(\tan \Theta\) and \(\cot \Theta\) from equation (2) ,(3) and (4) respectively from the expression  below

\(\frac{\cos \Theta – \frac{1}{\tan \Theta }}{2 \cot \Theta }\)

We get,

\(\frac{\cos \Theta – \frac{1}{\tan \Theta }}{2 \cot \Theta }\)= \(\frac{\frac{4}{5} – \frac{1}{3}}{2 \times \frac{4}{3}}\)

\(\frac{\cos \Theta – \frac{1}{\tan \Theta }}{2 \cot \Theta }\)= \(\frac{\frac{12}{15} – \frac{20}{15}}{\frac{8}{3}}\)

= \(\frac{\frac{ – 8}{15}}{\frac{8}{3}}\)

= \(\frac{- 1 }{5}\)

Therefore, \(\frac{\cos \Theta – \frac{1}{\tan \Theta }}{2 \cot \Theta }\)= \(\frac{- 1 }{5}\)

21.)   If \(\tan \Theta = \frac{24}{7}\), find that \(\sin \Theta + \cos \Theta\)

Sol.

Given:

\(\tan \Theta = \frac{24}{7}\) …. (1)

To find,

\(\sin \Theta + \cos \Theta\)

Now we know that \(\tan \Theta\) is defined as follows

\(\tan \Theta = \frac{Perpendicular\, side\, opposite\, to\, \angle \Theta }{Base\, side\, adjacent\, to \, \angle \Theta }\) …. (2)

Now by comparing equation (1) and (2)

We get,

Perpendicular side opposite to \(\angle \Theta\) = 24

Base side adjacent to \(\angle \Theta\) = 7

Therefore triangle representing \(\angle \Theta\) is as shown below

13

Side AC is unknown and can be found by using Pythagoras theorem

Therefore,

AC2= AB2 + BC2

Now by substituting the value of unknown sides from figure

We get,

AC2= 242 +72

AC = 576 + 49

AC= 625

Now by taking square root on both sides,

We get,

AC = 25

Therefore H hy

Hypotenuse side AC = 25 …. (3)

Now we know \(\sin \Theta\) is defined as follows

\(\sin \Theta = \frac{Perpendicular\, side\, opposite\, to \, \angle \Theta }{Hypotenuse}\)

Therefore from figure (a) and equation (3)

We get,

\(\sin \Theta = \frac{AB}{AC}\)

\(\sin \Theta = \frac{24}{25}\) …. (4)

Now we know that \(\cos \Theta\) is defined as follows

\(\cos \Theta = \frac{Base\, side\, adjacent\, to\, \angle \Theta }{Hypotenuse}\)

Therefore by substituting the value of \(\sin \Theta\) and \(\cos \Theta\) from equation (4) and (5) respectively, we get

\(\sin \Theta + \cos \Theta\) = \(\frac{24}{25} + \frac{7}{25}\)

\(\sin \Theta + \cos \Theta\) = \(\frac{31}{25}\)

             Hence,  \(\sin \Theta + \cos \Theta\) = \(\frac{31}{25}\)

 

22.)    If \(\sin \Theta = \frac{a}{b}\), find \(\sec \Theta + \tan \Theta\) in terms of a and b.

Sol.

             Given:

\(\sin \Theta = \frac{a}{b}\) …. (1)

To find: \(\sec \Theta + \tan \Theta\)

Now we know, \(\sin \Theta\) is defined as follows

\(\sin \Theta = \frac{Perpendicular\, side\, opposite\, to \, \angle \Theta }{Hypotenuse}\)       …. (2)

Now by comparing equation (1) and (2)

We get,

Perpendicular side opposite to \(\angle \Theta\) = a

Hypotenuse = b

Therefore triangle representing \(\angle \Theta\) is as shown below

14

Hence side BC is unknown

Now we find BC by applying Pythagoras theorem to right angled \(\Delta ABC\)

Therefore,

AC2 = AB2 +BC2

Now by substituting the value of sides AB and AC from figure (a)

We get,

b2 = a2 + BC2

Therefore,

BC2 = b2 – a2

Now by taking square root on both sides

We get,

BC= \(\sqrt{b^{2} – a^{2}}\)

Therefore,

Base side BC = \(\sqrt{b^{2} – a^{2}}\) …. (3)

Now we know \(\cos \Theta\) is defined as follows

\(\cos \Theta = \frac{Base\, side\, adjacent\, to\, \angle \Theta }{Hypotenuse}\)

Therefore from figure (a) and equation (3)

We get,

\(\cos \Theta = \frac{BC}{AC}\)

= \(\frac{\sqrt{b^{2} – a^{2}}}{b}\)

\(\cos \Theta = \frac{BC}{AC}\)

= \(\frac{\sqrt{b^{2} – a^{2}}}{b}\) …. (4)

Now we know, \(\sec \Theta = \frac{1}{\cos \Theta }\)

Therefore,

\(\sec \Theta =\frac{b}{\sqrt{b^{2}- a^{2}}}\) …. (5)

Now we know, \(\tan \Theta = \frac{\sin \Theta }{\cos \Theta }\)

Now by substituting the values from equation (1) and (3)

We get,

\(\tan \Theta =\frac{ \frac{a}{b}}{\frac{\sqrt{b^{2} – a^{2}}}{b}}\)

\(\tan \Theta =\frac{a}{\sqrt{b^{2}- a^{2}}}\)

Therefore,

\(\tan \Theta =\frac{a}{\sqrt{b^{2}- a^{2}}}\) …. (6)

Now we need to find \(\sec \Theta + \tan \Theta\)

Now by substituting the values of \(\sec \Theta\) and \(\tan \Theta\) from equation (5) and (6) respectively

We get,

\(\sec \Theta + \tan \Theta\) = \(\frac{b}{\sqrt{b^{2} – a^{2}}}+ \frac{a}{\sqrt{b^{2} – a^{2}}}\)

\(\sec \Theta + \tan \Theta\) = \(\frac{b + a}{\sqrt{b^{2} – a^{2}}}\) …. (7)

We get,

\(\sec \Theta + \tan \Theta\) = \(\frac{b + a}{\sqrt{b + a} – \sqrt{{b – a}}}\)

Now by substituting the value in above expression

We get,

\(\sec \Theta + \tan \Theta\)  = \(\frac{\sqrt{b + a}\times \sqrt{b + a}}{\sqrt{b + a} – \sqrt{{b – a}}}\)

Now, \(\sqrt{b + a}\) present in the numerator as well as denominator of above denominator of above expression gets cancels we get,

\(\sec \Theta + \tan \Theta=\frac{\sqrt{b + a}}{\sqrt{b -a}}\)

Square root is present in the numerator as well as denominator of above expression

Therefore we can place both numerator and denominator under a common square root sign

Therefore, \(\sec \Theta + \tan \Theta=\frac{\sqrt{b + a}}{\sqrt{b -a}}\)

 

23.)   If \(8\tan A = 15\) , find \(\sin A – \cos A\)

Sol.

             Given:

\(8\tan A = 15\)

Therefore,

\(\tan A = \frac{15}{8}\) …. (1)

To find:

\(\sin A – \cos A\)

Now we know tan A is defined as follows

\(\tan A = \frac{Perpendicular\, side\, opposite\, to \, \angle A}{Base \, side\, adjacent \, to\, \angle A}\) …. (2)

Now by comparing equation (1) and (2)

We get

Perpendicular side opposite to \(\angle A\) = 15

Base side adjacent to \(\angle A\) = 8

Therefore triangle representing angle A is as shown below

15

Side AC= is unknown and can be found by using Pythagoras theorem

Therefore,

AC2= AB2 + BC2

Now by substituting the value of known sides from figure (a)

We get,

AC2= 152 + 82

AC2 = 225 +64

AC = 289

Now by taking square root on both sides

We get,

AC = \(\sqrt{289}\)

AC = 17

Therefore Hypotenuse side AC=17 …. (3)

Now we know, sin A is defined as follows

\(\sin A = \frac{Perpendicular\, side\, opposite\, to \, \angle A}{Hypotenuse}\)

Therefore from figure (a) and equation (3)

We get,

\(\sin A= \frac{BC}{AC}\)

\(\sin A= \frac{15}{17}\) …. (4)

Now we know, cos A is defined as follows

\(\cos A = \frac{Base\, side\, adjacent\, to\, \angle A}{Hypotenuse}\)

Therefore from figure (a) and equation (3)

We get,

\(\cos A = \frac{AB}{AC}\)

\(\cos A = \frac{8}{17}\) …. (5)

Now we find the value of expression \(\sin A – \cos A\)

Therefore by substituting the value the value of \(\sin A\) and \(\cos A\) from equation (4) and (5) respectively , we get,

\(\sin A – \cos A = \frac{15}{17} – \frac{8}{17}\)

\(\sin A – \cos A = \frac{15 – 8}{17}\)

\(\sin A – \cos A = \frac{7}{17}\)

Hence, \(\sin A – \cos A = \frac{7}{17}\)

 

24.)    If \(\tan \Theta = \frac{20}{21}\), show that \(\frac{1 – \sin \Theta – \cos \Theta }{1+\sin \Theta +\cos \Theta } = \frac{3}{7}\)

Sol.
Given:

\(\tan \Theta = \frac{20}{21}\)

To show that \(\frac{1 – \sin \Theta  + \cos \Theta }{1+\sin \Theta +\cos \Theta } = \frac{3}{7}\)

Now we know that

\(\tan \Theta = \frac{Perpendicular\, side\, opposite\, to \, \angle \Theta}{Base \, side\, adjacent \, to\, \angle \Theta}\)

Therefore,

\(\tan \Theta = \frac{20}{21}\)

16
Side AC be the hypotenuse and can be found by applying Pythagoras theorem

Therefore,

AC2 = AB2 + BC2

AC2= 212 + 202

AC2 = 441 + 400

AC2 = 841

Now by taking square root on both sides

We get,

AC = \(\sqrt{841}\)

AC= 29

Therefore Hypotenuse side AC= 29

Now we know, \(\sin \Theta\) is defined as follows,

\(\sin A \Theta= \frac{Perpendicular\, side\, opposite\, to \, \angle \Theta}{Hypotenuse}\)

Therefore from figure and above equation

We get,

\(\sin \Theta = \frac{AB}{AC}\)

\(\sin \Theta = \frac{20}{29}\)

Now we know \(\cos \Theta\) is defined as follows

\(\cos \Theta = \frac{Base\, side\, adjacent\, to\, \angle \Theta}{Hypotenuse}\)

Therefore from figure and above equation

We get,

\(\cos \Theta = \frac{AB}{AC}\)

\(\cos \Theta = \frac{21}{29}\)

Now we need to find the value of expression \(\frac{1 – \sin \Theta  +  \cos \Theta }{1+\sin \Theta +\cos \Theta }\)

Therefore by substituting the value of \(\sin \Theta\) and \(\cos \Theta\)from above equations, we get

\(\frac{1 – \sin \Theta  +  \cos \Theta }{1+\sin \Theta +\cos \Theta }\) =

\(\frac{\frac{29 – 20 + 21}{29}}{\frac{70 }{29}}\)

Therefore after evaluating we get,

\(\frac{1 – \sin \Theta  +  \cos \Theta }{1+\sin \Theta +\cos \Theta }\)= \(\frac{3}{7}\)

Hence,

\(\frac{1 – \sin \Theta  +  \cos \Theta }{1+\sin \Theta +\cos \Theta }\)=

\(\frac{3}{7}\)

25.) If \(cosec A = 2\) , find \(\frac{1}{\tan A} + \frac{\sin A }{1 + \cos A}\)

Sol.

             Given:

\(cosec A = 2\)

To find \(\frac{1}{\tan A} + \frac{\sin A }{1 + \cos A}\)

Now cosec A = \(\frac{Hypotenuse}{Opposite side}\) = \(\frac{2}{1}\)

17

Here BC is the adjacent side,

By applying Pythagoras theorem,

AC2= AB2 + BC2

4 = 1 + BC2

BC2= 3

BC = \(\sqrt{3}\)

Now we know that

\(\sin A = \frac{1}{cosec A }\)

\(\sin A = \frac{1}{2}\) …. (1)

\(\tan A= \frac{AB}{BC}\)

\(\tan A= \frac{1}{\sqrt{3}}\) …. (2)

\(\cos A = \frac{BC}{AC}\)

\(\cos A = \frac{\sqrt{3}}{2}\) …. (3)

Substitute all the values of \(\sin A\) , \(\cos A\) and \(\tan A\) from the equations(1) ,(2) and (3) respectively

We get.

\(\frac{1}{\tan A} + \frac{\sin A }{1 + \cos A}\) = \(\frac{1}{\frac{1}{\sqrt{3}}} + \frac{\frac{1}{2}}{1 + \frac{\sqrt{3}}{2}}\)

= \(\sqrt{3} + \frac{1}{2 + \sqrt{3}}\)

=\(\frac{2(2 + \sqrt{3})}{2 + \sqrt{3}}\)

= 2

Hence,

\(\frac{1}{\tan A} + \frac{\sin A }{1 + \cos A}\) = 2

 

26.)   If \(\angle A\) and \(\angle B\) are acute angles such that cos A =cos B , then show that \(\angle A\)= \(\angle B\)

Sol.

Given:

\(\angle A\) and \(\angle B\) are acute angles

cos A = cos B such that \(\angle A\) = \(\angle B\)

Let us consider right angled triangle ACB

18

Now since cos A = cos B

Therefore

\(\frac{AC}{AB} = \frac{BC}{AB}\)

Now observe that denominator of above equality is same that is AB

Hence             \(\frac{AC}{AB} = \frac{BC}{AB}\) only when AC=BC

Therefore AC=BC

We know that when two sides of triangle are equal, then opposite of the sides are also

Equal.

Therefore

We can say that

Angle opposite to side AC = angle opposite to side BC

Therefore,

\(\angle B\) = \(\angle A\)

             Hence, \(\angle A\) = \(\angle B\)

 

27.)  In a \(\Delta ABC\) , right angled triangle at A, if tan C = \(\sqrt{3}\) , find the value of sin B cos C + cos B sin C.

Sol.

Given:

\(\Delta ABC\)

To find : sin B cos C + cos B sin C

The given a \(\Delta ABC\) is as shown in figure

19

Side BC is unknown and can be found using Pythagoras theorem,

Therefore,

BC2= AB2 +AC2

BC2= \(\sqrt{3}^{2}\) + 12
BC2 = 3 +1

BC2 = 4

Now by taking square root on both sides

We get,

BC = \(\sqrt{4}\)

BC= 2

Therefore Hypotenuse side BC= 2 …. (1)

Now, sin B = \(\frac{Perpendicular\, side\, opposite\, to\, \angle B}{Hypotenuse}\)

Therefore,

\(\sin B = \frac{AC}{BC}\)

Now by substituting the values from equation (1) and figure

We get,

sin B = \(\frac{1}{2}\) …. (2)

Now, cos B= \(\frac{base \, side\, adjacent\, to\, \angle B}{Hypotenuse}\)

Therefore,

cos B = \(\frac{AB}{BC}\)

Now substituting the value from equation

cos B= \(\frac{\sqrt{3}}{2}\) …. (3)

Similarly

sin C = \(\frac{\sqrt{3}}{2}\) …. (4)

Now by definition,

\(\tan C = \frac{sin C}{cos C}\)

So by evaluating

\(\cos C = \frac{1}{2}\) …. (5)

Now, by substituting the value of sinB, cosB,sin C and cosC from equation (2) ,(3) ,(4) and (5) respectively in sinB cosC + cosB sin C

sinB cosC + cosB sin C= \(\frac{1}{2}\times \frac{1}{2} + \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\)

= \(\frac{1}{4} + \frac{3}{4}\)

= 1

Hence,

sinB cosC + cosB sin C = 1

28.)   State whether the following are true or false. Justify your answer.

             (i)        The value od tan A is always less than 1.

             (ii)       sec A = \(\frac{12}{5}\) for some value of \(\angle A\).

            (iii) cos A is the abbreviation used for the cosecant of \(\angle A\).

           (iv) \(\sin \Theta = \frac{4}{3}\) for some angle \(\Theta\).

 

Sol.

(i) tan A \(<\) 1

Value of tan A at 45o i.e… tan  45 = 1

As value os A increases to 90o

Tan A becomes infinite

So given statement is false.

(ii) sec A = \(\frac{12}{5}\) for some value of angle if

M-I

sec A = 2.4

sec A > 1

So given statements is true.

M- II

For sec A = \(\frac{12}{5}\) we get adjacent side = 13

Subtending 9i at B.

So, given statement is true.

(iii) Cos A is the abbreviation used for cosecant of angle A.

The given statement is false.

As such cos A is the abbreviation used for cos of angle A , not as cosecant of angle A.

(iv) Cot A is the product of cot A and A

Given statement is false

\(∵\) cot A is a co-tangent of angle A and co-tangent of angle A = \(\frac{adjacent\, side }{Oposite\, side}\).

(v) \(\sin \Theta = \frac{4}{3}\) for some angle \(\Theta\).

Given statement is false

Since value of \(\sin \Theta\) is less than(or) equal to one.

Here value of \(\sin \Theta\) exceeds one,

So given statement is false.

29.)    If \(\sin \Theta = \frac{12}{13}\) find \(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\)

Sol.

Given: \(\sin \Theta = \frac{12}{13}\)

To Find:  \(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\)

As shown in figure

20

Here BC is the adjacent side,

By applying Pythagoras theorem,

AC2=AB2+BC2

169 = 144 +BC2

BC2= 169 – 144

BC2= 25

BC = 5

Now we know that,

\(cos \Theta = \frac{base \, side\, adjacent\, to\, \angle \Theta }{Hypotenuse}\)

\(\cos \Theta = \frac{BC}{AC}\)

\(\cos \Theta = \frac{5}{13}\)

We also know that,

\(\tan \Theta = \frac{\sin \Theta }{\cos \Theta }\)

Therefore, substituting the value of \(\sin \Theta\) and \(\cos \Theta\) from above equations

We get,

\(\tan \Theta = \frac{12}{5}\)

Now substitute all the values of \(\sin \Theta\) , \(\cos \Theta\) and \(\tan  \Theta\) from above equations in \(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\)

We get,

\(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\) = \(\frac{\left (\frac{12}{13} \right )^{2} – \left (\frac{5}{13} \right )^{2}}{2 \times \left (\frac{12}{13} \right )\times \left (\frac{5}{13} \right )} \times \frac{1}{\left (\frac{12}{5} \right )^{2}}\)

Therefore by further simplifying we get,

\(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\) = \(\frac{119}{169}\times \frac{169}{120}\times \frac{25}{144}\)

Therefore,

\(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\) = \(\frac{595}{3456}\)

Hence,

\(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\) = \(\frac{595}{3456}\)

 

30.)   If \(\cos \Theta = \frac{5}{13}\), find the value of \(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\)

Sol.

                Given: If \(\cos \Theta = \frac{5}{13}\)

To find:

The value of expression \(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\)

Now we know that

\(cos \Theta\)= \(\frac{base \, side\, adjacent\, to\, \angle \Theta}{Hypotenuse}\) …. (2)

Now when we compare equation (1) and (2)

We get,

Base side adjacent to \(\angle \Theta\) = 5

Hypotenuse = 13

Therefore, Triangle representing \(\angle \Theta\) is as shown below

21

Perpendicular side AB is unknown and it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

AC2= AB2 + BC2

Therefore by substituting the values ogf known sides,

AB2 = 132 – 52

AB2= 169 -25

AB2 = 144

AB = 12 …. (3)

Now we know from figure and equation,

\(\sin \Theta = \frac{12}{13}\) …. (4)

Now we know that,

\(tan\Theta = \frac{\sin \Theta }{\cos \Theta }\)

\(tan\Theta = \frac{12 }{5 }\) …. (5)

Now w substitute all the values from equation (1), (4) and (5) in the expression below,

\(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\)

Therefore

We get,

\(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\) = \(\frac{\left (\frac{12}{13} \right )^{2} – \left (\frac{5}{13} \right )^{2}}{2 \times \left (\frac{12}{13} \right )\times \left (\frac{5}{13} \right )} \times \frac{1}{\left (\frac{12}{5} \right )^{2}}\)

Therefore by further simplifying we get,

\(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\) = \(\frac{119}{169}\times \frac{169}{120}\times \frac{25}{144}\)

Therefore,

\(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\) = \(\frac{595}{3456}\)

Hence,

\(\frac{\sin ^{2}\Theta -\cos ^{2}\Theta}{2 \sin \Theta \cos \Theta } \times \frac{1}{\tan ^{2}\Theta }\) = \(\frac{595}{3456}\)

31.)   If sec A = \(\frac{17}{8}\) , verify that \(\frac{3 – 4 \sin ^{2}A}{4\cos ^{2}A – 3} = \frac{3 – \tan ^{2}A}{1 – 3 \tan ^{2}A}\)

Sol.

Given:   sec A = \(\frac{17}{8}\)

To verify: \(\frac{3 – 4 \sin ^{2}A}{4\cos ^{2}A – 3} = \frac{3 – \tan ^{2}A}{1 – 3 \tan ^{2}A}\)

Now we know that \(\cos A = \frac{1}{\sec A}\)

Now, by substituting the value of sec A

We get,

\(\cos A = \frac{8}{17}\)

Now we also know that,

\(\sin ^{2}A + \cos ^{2}A =1\)

Therefore

\(\sin ^{2}A =1 – \cos ^{2}A\)

= \(\left (\frac{8}{17} \right )^{2}\)

=\(\frac{225 }{289}\)

Now by taking square root on both sides,

We get,

\(\sin A= \frac{15}{17}\)

We also know that , \(\tan A = \frac{\sin A}{\cos A}\)

Now by substituting the value of all the terms ,

We get,

\(\tan A =\frac{15}{8}\)

Now from the expression of above equation which we want to prove:

L.H.S = \(\frac{3 – 4 \sin ^{2}A}{4\cos ^{2}A – 3}\)

Now by substituting the value of cos A ad sin A from equation (3) and (4)

We get,

L.H.S = \(\frac{3 – 4\frac{225}{289}}{4 – \frac{64}{289} – 3}\)

= \(\frac{867 – 900}{256 – 867}\)

= \(\frac{33}{611}\)

From expression

R.H.S = \(frac{3 – \tan ^{2}A}{1 – 3 \tan ^{2}A}\)

Now by substituting the value of tan A from above equation

We get,

R.H.S= \(\frac{3 – \left (\frac{15}{8} \right )^{2}}{1 – 3\left (\frac{15}{8} \right )^{2}}\)

= \(\frac{\frac{- 33}{64}}{\frac{-611}{64}}\)

= \(\frac{33}{611}\)

Therefore,

We can see that,

\(\frac{3 – 4 \sin ^{2}A}{4\cos ^{2}A – 3} = \frac{3 – \tan ^{2}A}{1 – 3 \tan ^{2}A}\)

 

32.)  If \(\sin \Theta = \frac{3}{4}\), prove that \(\sqrt{\frac{cosec^{2}\Theta – \cot ^{2}\Theta }{\sec ^{2} – 1}} = \frac{\sqrt{7}}{3}\)

Sol.

                Given: \(\sin \Theta = \frac{3}{4}\) …. (1)

To prove:

\(\sqrt{\frac{cosec^{2}\Theta – \cot ^{2}\Theta }{\sec ^{2} – 1}} = \frac{\sqrt{7}}{3}\) …. (2)

By definition,

sin A = \(\frac{Perpendicular\, side\, opposite\, to\, \angle A}{Hypotenuse}\) …. (3)

By comparing (1) and (3)

We get,

Perpendicular side = 3 and

Hypotenuse = 4

22

Side BC is unknown.

So we find BC by applying Pythagoras theorem to right angled \(\Delta ABC\)

Hence,

AC2 = AB2 +BC2

Now we substitute the value of perpendicular side (AB) and hypotenuse (AC) and get the base side (BC)

Therefore,

42= 32 +BC2

BC2 = 16 – 9

BC2= 7

BC = \(\sqrt{7}\)

Hence, Base side BC =\(\sqrt{7}\) …. (3)

Now cos A = \(\frac{BC}{AC}\)

\(\frac{\sqrt{7}}{4}\) …. (4)

Now , \(cosec A = \frac{1}{\sin A}\)

Therefore, from fig and equation (1)

\(cosec A = \frac{Hypotenuse}{Perpendicular}\)

\(cosec A = \frac{4}{3}\) …. (5)

Now, similarly

\(\sec A = \frac{4}{\sqrt{7}}\) …. (6)

Further we also know that

\(\cot A= \frac{\cos A}{\sin A}\)

Therefore by substituting th values from equation (1) and (4),

We get,

\(\cot A= \frac{\sqrt{7}}{3}\) …. (7)

Now by substituting the value of cosec A, sec A and cot A from the equations  (5), (6), and (7) in the L.H.S of expression (2)

\(\sqrt{\frac{cosec^{2}\Theta – \cot ^{2}\Theta }{\sec ^{2} – 1}}\) = \(\sqrt{\frac{\left (\frac{4}{3} \right )^{2} -\left ( \frac{\sqrt{7}}{3} \right )^{2}}{\left (\frac{4}{\sqrt{7}} \right )^{2} – 1}}\)

= \(\frac{\frac{16}{9}-\frac{7}{9}}{\frac{16}{7}-1}\)

= \(\frac{\sqrt{7}}{3}\)

Hence it is proved that,

    \(\sqrt{\frac{cosec^{2}\Theta – \cot ^{2}\Theta }{\sec ^{2} – 1}} = \frac{\sqrt{7}}{3}\)

 

33.)  If \(\sec A = \frac{17}{8}\) , verify that \(\frac{3 – 4\sin ^{2}A}{4\cos ^{2}A – 3}= \frac{3- \tan ^{2}A}{1 – 3\tan ^{2}A}\)

Sol.

                Given: \(\sec A = \frac{17}{8}\) …. (1)

To verify:

\(\frac{3 – 4\sin ^{2}A}{4\cos ^{2}A – 3}= \frac{3- \tan ^{2}A}{1 – 3\tan ^{2}A}\) …. (2)

Now we know that sec A = \(\frac{1}{cos A}\)

Therefore \(\cos A= \frac{1}{sec A}\)

We get,

\(\cos A= \frac{8}{17}\) …. (3)

Similarly we can also get,

sin A= \(\sin A= \frac{15}{17}\) …. (4)

An also we know that \(\tan A= \frac{sin A}{cos A}\)

\(\tan A= \frac{15}{8}\) …. (5)

Now from the expression of equation (2)

L.H.S: \(\frac{3 – 4\sin ^{2}A}{4\cos ^{2}A – 3\)

Now by substituting the value of cos A and sin A from equation (3) and (4)

We get,

L.H.S = \(\frac{3 – 4\left (\frac{15}{17} \right )^{2}}{4\left (\frac{8}{17} \right )^{2} – 3}\)

= \(\frac{\frac{867 -900}{289}}{\frac{256 – 867}{289}}\)

=\(\frac{33}{611}\) …. (6)

R.H.S = \(\frac{3- \tan ^{2}A}{1 – 3\tan ^{2}A}\)

Now by substituting the value of tan A from equation (5)

We get,

R.H.S=\(\frac{3 – \left (\frac{15}{18} \right )^{2}}{1 – 3\left (\frac{15}{8} \right )^{2}}\)

\(\frac{\frac{- 33}{64}}{\frac{-611}{64}}\)

=\(\frac{33}{611}\) …. (7)

Now by comparing equation (6) and (7)

We get,

\(\frac{3 – 4\sin ^{2}A}{4\cos ^{2}A – 3}= \frac{3- \tan ^{2}A}{1 – 3\tan ^{2}A}\)

 

34.) If \(\cot \Theta = \frac{3}{4}\), prove that \(\frac{\sec \Theta – cosec\Theta }{\sec \Theta + cosec\Theta } = \frac{1}{\sqrt{7}}\)    

Sol.

Given: \(\cot \Theta = \frac{3}{4}\)

Prove that: \(\frac{\sec \Theta – cosec\Theta }{\sec \Theta + cosec\Theta } = \frac{1}{\sqrt{7}}\)

Now we know that

\(\frac{\sec \Theta – cosec\Theta }{\sec \Theta + cosec\Theta } = \frac{1}{\sqrt{7}}\)

23

Here AC is the hypotenuse and we can find that by applying Pythagoras theorem

AC2= AB2 +BC2

AC2 = 16 +9

AC2= 25

AC = 5

Similarly

\(\sec \Theta = \frac{AC}{BC}\)

\(\sec \Theta = \frac{5}{3}\)

\( cosec = \frac{AC}{AB}\)

\( cosec = \frac{5}{4}\)

Now on substituting the values in equations we get,

\(\frac{\sec \Theta – cosec\Theta }{\sec \Theta + cosec\Theta } =\frac{1}{\sqrt{7}}\)

Therefore,

\(\frac{\sec \Theta – cosec\Theta }{\sec \Theta + cosec\Theta } =\frac{1}{\sqrt{7}}\)    

 

35.) If \(3 \cos \Theta – 4\sin \Theta = 2\cos \Theta + \sin \Theta\) ,find \(\tan \Theta\)           

Sol.

Given: \(3 \cos \Theta – 4\sin \Theta = 2\cos \Theta + \sin \Theta\)

To find: \(\tan \Theta\)

We can write this as:

\(3 \cos \Theta – 4\sin \Theta = 2\cos \Theta + \sin \Theta\)

\(\cos \Theta = 5 \sin \Theta\)

Dividing both the sides by \(\cos \Theta\) ,

We get,

\(\frac{\cos \Theta }{\cos \Theta }=\frac{ 5 \sin \Theta }{\cos \Theta }\)

\(1 = 5\tan \Theta\)

\(\tan \Theta = 1\)

Hence,

    \(\tan \Theta = 1\)

 

36.)    If \(\angle A\) and \(\angle P\) are acute angles such that tan A = tan P, then show \(\angle A = \angle P\)

Sol.

Given: A and P are acute angles tan A =tan P

Prove that: \(\angle A = \angle P\)

Let us consider right angled triangle ACP

24

We know \(\tan \Theta = \frac{opposite side}{adjacent side}\)

tan A =\(\frac{PC}{AC}\)

tan P =\(\frac{AC}{PC}\)

\(∴\) tan A =tan P

\(\frac{PC}{AC}= \frac{AC}{PC}\)

PC =AC [\(∵\)Angle opposite to equal sides are equal]

\(\angle A = \angle P\)<