# 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

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

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

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

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

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

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

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

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

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

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 = $\sqrt{10}$

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

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:

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,

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

$\cos A = \frac{AB}{AC}$

$\cos A = \frac{24}{25}$

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

(ii)  The given triangle is below:

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,

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

$\cos A = \frac{BC}{AC}$

$\cos A = \frac{7}{25}$

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

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

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

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

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,

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

$\tan A = \frac{BC}{AB}$

$\tan A = \frac{9}{40}$

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

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

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

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

$\sec A= \frac{17}{8}$

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

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

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

$\sec R = \frac{PR}{RQ}$

$\sec R = \frac{5}{3}$

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

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

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,

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

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

$\frac{49}{64}$

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

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.

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

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,

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

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

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

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

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

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

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

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

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

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,

$\tan \Theta = \frac{20}{21}$

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 = $\frac{Hypotenuse}{Opposite side}$ = $\frac{2}{1}$

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

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

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

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

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

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

$\frac{\sec \Theta – cosec\Theta }{\sec \Theta + cosec\Theta } = \frac{1}{\sqrt{7}}$

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

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]

$\angle A = \angle P$<