RD Sharma Solutions Class 10 Trigonometric Ratios Exercise 5.1

RD Sharma Class 10 Solutions Chapter 5 Ex 5.1 PDF Free 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.

Solution:

(i) sin A = 2/3

Given here;

sin A = 2/3 ……..….. (1)

As we know, by sin definition;

sin A Perpendicular/Hypotenuse = 2/3 ….(2)

On Comparing eq. (1) and (2), we get;

Perpendicular = 2

And, Hypotenuse = 3

1

Now, using Pythagoras theorem,

AC2 = AB2 + BC2

Putting 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= √5

Hence, Base = √5

Since, cos A = Base/Hypotenuse

Thus, cos A = √5/3

Since, cosec A = 1/sin A

So, cosec A =Hypotenuse/Perpendicular

Thus, cosec A =3/2

Since, sec A = Hypotenuse/Base

So, sec A = 3/√5

Since tan A = Perpendicular/Base

So, tan A =  2/√5

Since, cot A = Base/Perpendicular

So, cot A = √5/2

 

(ii) cos A = 4/5

Given here:    cos A = 4/5 …….…. (1)

As we know,

cos A = Base/Hypotenuse …. (2)

On Comparing eq. (1) and (2), we get;

Base =4 and

Hypotenuse = 5

1

Now, using Pythagoras theorem,

AC2 = AB2 + BC2

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

52 = 42+ BC2

BC2 = 52 – 42

BC2= 25 – 16

BC2 = 9

BC= 3

Hence, Perpendicular = 3

Since, sin A = Perpendicular/Hypotenuse = 2/3

∴ sin A= 3/5

Since, cosec A = 1/sin A

So, cosec A=  1/sinA

Or cosec A =Hypotenuse/Perependicular

∴ cosec A = 5/3

Since, sec A =1/cos A

So, sec A =Hypotenuse/Base

Therefore, sec A = 5/4

Since, tan A = Perpendicular/Base

∴ tan A = 3/4

Since, cot A = 1/tan A

So, cot A = Base/Perpendicular

∴cot A = 4/3

 

(iii) tan θ = 11

Given here;  tan θ = 11…..…. (1)

As we know,

tan θ = Perpendicular/Base…. (2)

On Comparing eq. (1) and (2), we get;

Base= 1 and Perpendicular = 5

2

Now, using Pythagoras theorem,

AC2 = AB2 + BC2

Putting the value of base (AB) and perpendicular (BC) to get hypotenuse(AC), we get;

AC2 = 12 + 112

AC2 = 1 + 121

AC2= 122

AC= √122

Now, sin = Perpendicular/Hypotenuse

sin θ = 11/√122

Since, cosec θ = 1/sin θ

So, cosec θ = √122/11

Since, cos θ = Base/Hypotenuse

cos θ = 1/√122

Since, sec θ = 1/cos θ

∴ sec θ = √122/1 = √122

Since, cot θ  = 1/tan θ

cot θ = 1/11

 

(iv) sin θ = 11/15

Given here;  sin θ = 11/15 ………. (1)

As we know,

 sin θ = Perpendicular/Hypotenuse …. (2)

On Comparing eq. (1) and (2), we get;

Perpendicular = 11 and Hypotenuse= 15

Now, using Pythagoras theorem,

AC2 = AB2 + BC2

1

Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base (AB), we get;

152 = AB2 +112

AB2 = 152 – 112

AB2= 225 – 121

AB2 = 104

AB = √104

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

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

AB= 2√26

Hence, Base = 2√26

Since, cos θ = Base/Hypotenuse

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

Since, cosec θ = 1/sin θ

cosec θ = 15/11

Since, secθ = Hypotenuse/Base

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

Since, tan θ = Perpendicular/Base

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

Since, cot θ = Base/Perpendicular

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

 

(v) tanj α = 5/12

Given here;  tan α = 5/12 …. (1)

As we know,

tan α = Perpendicular/Base…. (2)

On Comparing eq. (1) and (2), we get;

Base= 12 and Perpendicular side = 5

1

Now, using Pythagoras theorem,

AC2 = AB2 + BC2

Putting the value of base (AB) and the perpendicular (BC) to get hypotenuse (AC), we get;

AC2 = 122 + 52

AC2 = 144 + 25

AC2= 169

AC= 13

Hence, Hypotenuse = 13

Since, sin α  = Perpendicular/Hypotenuse

∴ sin α = 5/13

Since, cosec α  = Hypotenuse/Perpendicular

∴ cosec α = 13/5

Since, cos α = Base/Hypotenuse

cos α = 12/13

Since, sec α =1/cos α

∴ sec α = 13/12

Since, tan α = sin α/cos α

∴ tan α=5/12

Since, cot α = 1/tan α

cot α =12/5

 

(vi) sin θ = √3/2

Given here; sin θ = √3/2 …………. (1)

As we know,

sin θ = Perpendicular/Hypotenuse….(2)

On Comparing eq. (1) and (2), we get;

Perpendicular = √3 and Hypotenuse = 2

1

Now, using Pythagoras theorem,

AC2 = AB2 + BC2

Putting the value of perpendicular (BC) and hypotenuse (AC) and get the base (AB), we get;

22 = AB2 + (√3)2

AB2 = 22 – (√3)2

AB2 = 4 – 3

AB2 = 1

AB= 1

Thus, Base = 1

Since, cos θ = Base/Hypotenuse

cos θ = 1/2

Since, cosec θ  = 1/sin θ

Or cosec θ= Hypotenuse/Perpendicualar

∴ cosec θ =2/√3

Since, sec θ = Hypotenuse/Base

∴ sec θ = 2/1

Since, tan θ = Perpendicula/Base

tan θ = √3/1

Now, cot θ = Base/Perpendicular

cot θ = 1/√3

 

(vii) cos θ = 7/25

Given here; cos θ = 7/25 ……….. (1)

As we know,

cos θ = Base/Hypotenuse

On Comparing eq. (1) and (2), we get;

Base = 7 and Hypotenuse = 25

1

Now, using Pythagoras theorem,

AC2= AB2 + BC2

Putting the value of base (AB) and hypotenuse (AC) to get the perpendicular (BC)

252 = 72 +BC2

BC2 = 252 – 72

BC2 = 625 – 49

BC = 576

BC= √576

BC= 24

Hence, Perpendicular side = 24

Since, sin θ = perpendicular/Hypotenuse

∴ sin θ = 24/25

Since, cosec θ = 1/sin θ

So, cosec θ= Hypotenuse/Perpendicualar

∴ cosec θ = 25/24

Since, sec θ = 1/cosec θ

So, sec θ = Hypotenuse/Base

∴ sec θ = 25/7

Since, tan θ = Perpendicular/Base

tan θ = 24/7

Now, cot = 1/tan θ

So, cot θ = Base/Perpendicular

∴ cot θ = 7/24

 

(viii) tan θ = 8/15

Given here; tan θ = 8/15 …………. (1)

As we know,

tan θ = Perpendicular/Base …. (2)

On Comparing eq. (1) and (2), we get;

Base= 15 and Perpendicular = 8

1

Now, using Pythagoras theorem, we get;

AC2= 152 + 82

AC2= 225 + 64

AC2 = 289

AC = √289

AC= 17

Hence, Hypotenuse = 17

Since, sin θ = perpendicular/Hypotenuse

 sin θ = 8/17

Since, cosec θ = 1/sin θ

So, cosec θ= Hypotenuse/Perpendicualar

∴ cosec θ = 17/8

Since, cos θ = Base/Hypotenuse

cos θ = 15/17

Since, sec θ= 1/cos θ 

So, sec θ = Hypotenuse/Base

∴ sec θ = 17/15

Since, cot θ = 1/tan θ 

So, cot θ = Base/Perpendicular

∴ cot θ = 15/8

 

(ix) cot θ = 12/5

Given here;  cot θ = 12/5 …………. (1)

As we know,

cot θ = 1/tan θ

cot θ = Base/Perpendicular ……. (2)

On Comparing eq. (1) and (2), we get;

Base = 12 and Perpendicular side = 5

1

Now, using Pythagoras theorem, we get;

AC2= AB2 + BC2

Putting the value of base (AB) and perpendicular (BC) to get the hypotenuse (AC);

AC2 = 122 + 52

AC2= 144 + 25

AC2 = 169

AC = √169

AC = 13

Hence, Hypotenuse = 13

Since, sin θ = perpendicular/Hypotenuse

sin θ= 5/13

Since, cosec θ = 1/sin θ 

cosec θ= Hypotenuse/Perpendicualar

cosec θ = 13/5

Since, cos θ = Base/Hypotenuse

cos θ = 12/13

Since, sec θ = 1/cosθ 

So, sec θ = Hypotenuse/Base

∴ sec θ = 13/12

Since, tanθ = 1/cot θ 

So, tan θ = Perpendicular/Base

∴ tan θ = 5/12

 

(x) sec θ= 13/5

Given here;  sec θ = 13/5…….… (1)

As we know,

sec θ = Hypotenuse/Base…………. (2)

On Comparing eq. (1) and (2), we get;

Base=5 and Hypotenuse = 13

1

Now, using Pythagoras theorem,

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

132 = 52 + BC2

BC2 = 132 – 52

BC2=169 – 25

BC2= 144

BC= √144

BC = 12

Hence, Perpendicular = 12

Since, sin θ = perpendicular/Hypotenuse

∴ sin θ= 12/13

Since, cosec θ= 1/ sin θ

So, cosec θ= Hypotenuse/Perpendicualar

∴ cosec θ = 13/12

Since, cos θ= 1/sec θ

So,  cos θ = Base/Hypotenuse

∴cos θ = 5/13

Since, tan θ = Perpendicular/Base

tan θ = 12/5

Since, cot θ = 1/tan θ

So, cot θ = Base/Perpendicular

∴ cot θ = 5/12

 

(xi) cosec θ= √10

Given here; cosec θ = √10/1   ……..… (1)

As we know,

cosecθ= 1/sin θ  …….….(2)

cosec θ= Hypotenuse/Perpendicualar

On comparing eq.(1) and(2), we get;

Perpendicular side= 1 and Hypotenuse = √10

1

Now, using Pythagoras theorem,

AC2 = AB2 + BC2

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

(√10)2 = AB2 + 12

AB2= (√10)2 – 12

AB2= 10 – 1

AB = √9

AB = 3

So, Base side = 3

Since, sin θ = perpendicular/Hypotenuse

sin θ = 1/√10

Since,  cos θ = Base/Hypotenuse

cos θ = 3/√10

Since, sec θ = 1/cos θ

sec θ = Hypotenuse/Base

sec θ = √10/3

Since, tan θ = Perpendicular/Base

tan θ = 1/3

Since, cot θ = 1/tan θ 

∴ cot θ = 3/1

or cot θ = 3

 

(xii) cos θ = 12/15

Given here; cos θ = 12/15 ………. (1)

As we know,

cos θ = Base/Hypotenuse……… (2)

On Comparing eq. (1) and (2), we get;

Base=12 and Hypotenuse = 15

1

Now, using Pythagoras theorem, we get;

AC2 = AB2+ BC2

Puttingthe value of base (AB) and hypotenuse (AC) to get the perpendicular (BC);

152 = 122 + BC2

BC2 = 152 – 122

BC2 = 225 – 144

BC 2= 81

BC = √81

BC = 9

So, Perpendicular = 9

Since, sin θ = perpendicular/Hypotenuse

 sin θ = 9/15 = 3/5

Since, cosec θ = 1/sin θ

So, cosec θ= Hypotenuse/Perpendicualar

∴ cosec θ= 15/9 = 5/3

Since, sec θ = 1/cos θ

So, sec θ = Hypotenuse/Base

∴ sec θ = 15/12 = 5/4

Since, tan θ = Perpendicular/Base

tan θ = 9/12 = 3/4

Since, cot θ = 1/tan θ

So, cot θ = Base/Perpendicular

∴ cot θ = 12/9 = 4/3

 

2.)   In a △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 here; In △ABC , AB= 24 cm

BC = 7cm

∠ABC = 90o

To find: sin A, cos A

First, we have to find the Hypotenuse side.

So, using Pythagoras theorem, we get;

AC2 = AB2 + BC2

AC2 = 242 + 72

AC2 = 576 + 49

AC2= 625

AC = √625

AC= 25

Hence, Hypotenuse = 25

As we know,

sin A = Perpendicular side opposite to angle A/Hypotenuse

sin A = BC/AC

sin A = 7/25

As we know,

cos A = Base side adjacent to angle A/Hypotenuse

cos A = AB/AC

cos A = 24/25

 

(ii)  Given, a triangle in below figure:

3

Given here;  In △ABC , AB= 24 cm and BC = 7cm

∠ABC = 90o

To find: sin C, cos C ?

First, we have to find the Hypotenuse side.

Now, using Pythagoras theorem, we get;

AC2 = AB2 + BC2

AC2 = 242 + 72

AC2 = 576 + 49

AC2= 625

AC = √625

AC= 25

Hence, Hypotenuse = 25

As we know,

sin C = Perpendicular side opposite to angle C/Hypotenuse

sin C = AB/AC

sin C = 24/25

As we know,

cos C = Base side adjacent to angle C/Hypotenuse

cos A = BC/AC

cos A = 7/25

 

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

  4                   

               

Solution: To find, tan P, cot R?

In the given right-angled △PQR, length of base side OR is unknown to us.

Therefore , by using Pythagoras theorem in △PQR, we get;

PR2 = PQ2 + QR2

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

132= 122 + QR2

QR2 = 132 – 122

QR2 = 169 – 144

QR2= 25

QR = √25

As we know that ,

tan P = Perpendicular side opposite to P/Base side adjacent to angle P

tan P = QR/PQ 

tan P = 5/12 ………. (1)

As we know,

cot R= Base/Perpendicular

cot R=QR/PQ

cot R=5/12 …. (2)

If we compare equation (1) ad (2), we can see that R.H.S of both the equation is equal.

Therefore, L.H.S of both equations will also be equal;

tan P = cot R

Yes , tan P =cot R = 5/12

 

4.)  If sin A = 9/41, Compute cos A and tan A.

Solution:

Given here; sin A= 9/41 …………. (1)

To find: cos A, tan A ?

As we know,

sin A= Perpendicular/Hypotenuse……………(2)

On Comparing eq. (1) and (2), we get;

Perpendicular side = 9 and Hypotenuse = 41

Now, using the perpendicular and hypotenuse, we can construct △ABC as shown below;

5

Length of base AB is unknown to us in right-angled △ABC,

Therefore, by using Pythagoras theorem in △ABC, we get;

AC2 = AB2 + BC2

412 = AB2 + 92

AB2 = 412 – 92

AB2 = 168 – 81

AB= 1600

AB = √1600

AB = 40

Hence, Base of triangle ABC, AB= 40

As we know,

cos A = Base/Hypotenuse

cos A =AB/AC

cos A =40/41

As we know,

tan A = Perpendicular/Base

tan A = BC/AB

tan A = 9/40

 

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

Solution;

Given here; 15cot A = 8

To find: sin A, sec A ?

Since 15 cot A =8

Dividing both the sides by 15, we get;

cot A = 8/15 …….(1)

As we know,

cot A = 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)

On comparing equation (1) and (2), we get;

Base side adjacent to ∠A = 8

Perpendicular side opposite to ∠A = 15

△ABC can be drawn below based on above information.

6

The hypotenuse is still unknown.

So, by using Pythagoras theorem to △ABC, we get;

AC2 = AB2 +BC2

Putting values of sides from the above figure

AC2 = 82 + 152

AC2 = 64 + 225

AC2 = 289

AC = \(\sqrt{289}\)

AC = 17

Therefore, hypotenuse =17

As we know,

sin A=Perpendicular/Hypotenuse

Therefore, sin A= BC/AC

Putting values of sides from the above figure;

sin A= 15/17

As we know,

sec A= 1/cos A

Hence,

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

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

Putting values of sides from the above figure;

sec A= 17/8

 

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

Solution: Given here;

△PQR is right-angled at Q.

PQ= 4cm

RQ= 3cm

To find: sin P, sin R , sec P , sec R ?

Given △PQR is as shown below;

7

Hypotenuse PR is unknown to us

By using Pythagoras theorem to △PQR, we get;

PR2 = PQ2 +RQ2

Putting values of sides from the above figure;

PR2 = 42 +32

PR2 = 16 + 9

PR2 = 25

PR = \(\sqrt{25}\)

PR = 5

Hence, Hypotenuse =5

As we know,

sin P =Perpendicular side opposite to angle P/Hypotenuse

sin P =RQ/PR

Putting values of sides from the above figure;

sin P = 3/5

As we know,

sin R = Perpendicular side opposite to angle R/Hypotenuse

sin R = PQ/PR

Putting the values of sides from above figure;

sin R = 4/5

As we know,

sec P=1/cos P

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

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

Putting values of sides from the above figure, we get;

sec P = PR/PQ

sec P = 5/4

As we know,

sec R = 1/cos R

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

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

Putting values of sides from the above figure;

sec R = PR/RQ

sec R = 5/3

Answer:  sin R = 4/5, sec P = 5/4, sec R = 5/3

 

7.)      If cot θ= 7/8, evaluate

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

       (ii)  cot2θ

Solution.

Given here; cot θ = 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,

(a + b)(a – b) = a2 – b2

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

We will get,

(1 + sin θ ) × (1 – sin θ ) = 1 – sin2 θ . . . . . (2) (Where, a = 1 and b = sin θ)

In the same way,

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

We will get,

(1 + cos θ )(1 – cos θ )= 12 – cos2 θ… (Where a=1 and b= cosθ)

(1 + cos θ )(1 – cos θ )= 1 – cos2 θ … (Where a=1 and b= cosθ)

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

Hence,

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

Since,

cos2θ  + sin2θ = 1

Therefore,

cos2θ  = 1 – sin2θ

Also, sin2θ = 1 – cos2θ

Putting the value of 1 – sin2θ and 1 – cos2θ 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, cosθ/sinθ = cot θ

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

Since, it is given here, cot θ = 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 here;cot θ = 7/8

To find: cot2θ

cot θ = 7/8

On squaring both the sides,

We get,

(cot θ)2= (7/8)2

(cot θ)2=49/64. Answer

 

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

Solution:

Given here; 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

Divide by 3 on both the sides of above equation.

cot A = 4/3…. (1)

As we know,

cot A = 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)

On comparing equation (1) and (2), we get;

Base side adjacent to ∠A = 4

Perpendicular side opposite to ∠A = 3

Hence △ABC is as shown in figure below;

8

In △ABC , Hypotenuse side is unknown

Hence, it can be found by using Pythagoras theorem

Therefore, by using Pythagoras theorem in △ABC, we get;

AC2= AB2 +BC2

Putting the values of sides from the above figure, we get;

AC2 =42 + 32

AC2 = 16 +9

AC2 = 25

AC = \(\sqrt{25}\)

AC = 5

Hence, hypotenuse= 5

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

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

As we know,

tan A = 1/cot A

Putting the value of cot A from equation (1), we get;

tan A  = 1/4

tan A = 3/4  ……… (3)

As we know,

sin A = Perperdicular side/Hypotenuse side

So, sin A = 3/5 ……..(4)

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

cos A = AB/AC 

Putting the values of sides from the above figure, we get;

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

Putting value of tan A from equation (3)

We get,

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

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

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

Take L.C.M of both numerator and denominator;

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

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

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

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)

On Comparing eq. (6) and (7);

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

Hence, proved.

 

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

Solution.

Given here;

tan θ =a/b ….…. (1)

Now, we know that tan θ  = sin θ/cos θ

Therefore, we can write equation (1) as;

sin θ/cos θ = a/b

Now, by using invertendo;

cos θ/sin θ = b/a

Now by using Componendo – dividendo both;

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

Hence,

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

 

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

Solution.

Given here; If tan θ =4

Therefore,

tan θ = 4/3……… (1)

Now, we know, tan θ = sin θ/cos θ

Therefore, equation (1) can be written as;

sin θ/cos θ = 4/3 ….(2)

Now, by using Invertendo to equation (2);

cos θ /sin θ = 3/4…. (3)

On multiplying by 4 both the sides;

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

Hence, we get;

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

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

On multiplying by 2 both the sides of equation (3);

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

Now by using 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)

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

Hence, we get;

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

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

Hence, 4cos θ – sin θ = 4

 

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

Solution: Given here;

3 cot θ= 2

Therefore,

\(\cot \Theta = \frac{2}{3}\) …. (1)

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

Thus, equation (1) becomes;

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

Now , by using invertendo to equation (2)

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

On multiplying by \(\frac{4}{3}\) both the sides;

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

On solving the above expression, we get;

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

Now by using invertendo-dividendo method in above equation;

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

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

On multiplying by \(\frac{2}{6}\) both the sides of equation (3);

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

On solving the above expression, we get;

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

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

Now by using componendo in above equation;

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

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

On dividing equation (4) by equation (5);

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

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

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

Therefore, on L.H.S (3 sinθ) 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 as common factor, from both the sides;

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

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

 

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

Solution: Given here;

tan θ= a/b …. (1)

We know, tan θ = sin θ/cos θ

Thus, equation (1) becomes;

sin θ/cos θ =a/b….(2)

On multiplying by \(\frac{a}{b}\) both the sides of equation (2);

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

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

Now, by using dividendo in above equation (3);

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

Now by using componendo in equation (3);

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

On dividing equation (4) by equation (5), we get;

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

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

On solving the above expression, we get;

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

Hence, proved.

 

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

 Solution.

Given here;

sec θ= 13/5

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

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

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

\(\cos \Theta = \frac{5}{13}\) …. (1)

We already know;

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

On comparing equation (1) and(2);

Base side adjacent to ∠θ = 5 And Hypotenuse =13

9

From the above figure, we find;

Base side BC = 5

Hypotenuse AC = 13

Side AB, perpendicular, is unknown.

Therefore, by using Pythagoras theorem;

AC2 = AB2 + BC2

Now, by putting the values of base and hypotenuse;

132 = AB2 + 52

Or,

AB2= 132 – 52

AB2= 169 – 25

AB2 = 144

AB = \(\sqrt{144}\)

So, AB = 12 …. (3)

We know;

sin θ= AB/AC

sin θ = 12/13 …. (4)

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

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

Putting the value cosθ and  sinθ and from equation (1) and (4) respectively;

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

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= 9/3

L.H.S= 3

Hence, proved.

 

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

Solution.

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

To prove: \(\sin \Theta (1 – \tan \Theta )= \frac{35}{156}\)

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

On comparing equation (1) and (2);

Base side adjacent to ∠θ = 12

And

Hypotenuse = 13

10

From the above figure, we get;

Base side BC= 12

Hypotenuse AC= 13

Side AB, perpendicular, is unknown and therefore by using Pythagoras theorem;

AC2= AB2 + BC2

Therefore, by Putting the values of known base and hypotenuse, we get;

132= AB2 + 122

Or

AB2= 132– 122

AB2= 169 – 144

AB =25

AB= \(\sqrt{25}\)

Therefore,

AB = 5 …. (3)

As we know;

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

From the figure, we can see;

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

Therefore,

\(\sin \Theta = \frac{5}{12}\) …. (5)

Now taking L.H.S;

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

Putting the value of sinθ and tanθ from equation (4) and (5), we get;

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

Taking L.C.M of the denominator;

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 expanding the above equation and solving it;

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

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

Hence, proved.

 

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

Solution.

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

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

We know already, \(\cot \Theta = \frac{1}{\tan \Theta }\)

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

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

Therefore,

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

On comparing Equation (1) and (2), we get;

Base side adjacent to ∠θ = 1

Perpendicular side opposite to ∠θ = √3

Hence, triangle representing side √3 is as shown below

11

Therefore, by Putting the values of base and perpendicular sides, we get;

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

Or

AC2 = 3 + 1

AC2= 4

AC= √4

Hence the value of hypotenuse,

AC = 2 …….. (3)

As we know already;

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

From the figure, we can say;

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

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

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

As we know already;

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

From the figure, we have;

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

Hence, from the figure and equation (3),

\(\cos \Theta = \frac{1}{2}\) …. (5)

Now Taking L.H.S.,

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

Putting the value 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 the L.C.M, we will get;

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

Hence,

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

Hence, proved.

 

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 here; \(\tan \Theta = \frac{1}{\sqrt{7}}\) …. (1)

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

Now, we know that

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

On comparing equation (1) and (2), we get;

Perpendicular side opposite to ∠θ = 1

Base side adjacent to ∠θ= \(\sqrt{7}\)

Now, the right-angled triangle representing ∠θ is shown below;

12

Hypotenuse side AC is unknown.

Thus, by using Pythagoras theorem, we get;

AC2= AB2 + BC2

Therefore by putting the values of base and perpendicular side, we get;

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

Or

AC 2 = 1 +7

AC2 = 8

AC = \(\sqrt{8}\)

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

Thus,

AC = \(2\sqrt{2}\) …. (3)

As 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)

As we know that;

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

So, from equation (4), we get;

cosec \(\Theta = 2\sqrt{2}\) …. (5)

As we know that;

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

So, from the figure, we get;

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

Again, from the figure and equation (3), we get;

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

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

Now, 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, taking the L.H.S;

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

Putting the value of cosec\(\Theta\) and secθ 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= \(\frac{\left (8 \right )- \left (\frac{8}{7} \right )}{\left (8 \right ) + \left (\frac{8}{7} \right )}\)

Hence,

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

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

Hence,

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

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

Hence,

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

 

 

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

Solution:

Given here; sec θ = 5/4 …. (1)

We need to find the value of :

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

We know that, sec θ = 1/cos θ, thus;

cos θ = 1/sec θ 

So, from equation (1), we can write;

cos θ = 1/5

cos θ = 4/5 …. (2)

Also, we know that, cos2θ  + sin2θ = 1;

Therefore, we can say;

sin2θ = 1 – cos2θ

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

Putting the value of cos θ, from equation (2), we get;

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

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

= 9/25

= 3/5

So, sinθ = 3/5……….. (3)

We know from trigonometric identities;

sec2θ = 1 + tan2θ

So, we can write;

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

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

Therefore,

tanθ =3/4 …. (4)

Since, cotθ = 1/tanθ

Thus, from equation (4), we can write as;

cotθ = 4/3 …. (5)

Putting the value of cos θ, cot θ and tan θ from the equation (2),(3),(4) and (5) in the expression below, we get;

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

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

=12/7 = Answer

 

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

Solution:

Given here; sinθ = 12/13  …. (1)

We need to find the value of : \(\frac{2\sin \Theta \cos \Theta }{\cos ^{2}\Theta – \sin ^{2}\Theta }\)

Now, we know from trigonometric identity;

cosec2θ = 1 + tan2θ

Therefore, by Putting the value of tan θ from equation (1), we get here;

cosec2θ = 1 +(12/13)2

cosec2θ = 1 + 122/132

cosec2θ = 1 + 144/169

Taking L.C.M of the denominator we get;

cosec2θ = (196+144)/163

cosec2θ = 313/169

Or

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

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

Therefore

cosec θ = √313/13 …. (2)

As we already know;

cosecθ = 1/sinθ

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

So,the value of;

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

Now, we know from the trigonometric identity;

cos2θ  + sin2θ = 1

or we can write it as;

cos2θ = 1 – sin2θ

Now by Putting the value of sinθ from equation (3), we get;

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

=\(1 – \frac{169}{313}\)

Now by taking L.C.M of R.H.S, we get;

cos2θ  = 144/313

Or we can write as;

cosθ = 12/√313 ……….(4)

Putting the value of sinθ and cosθ from equation (3) and (4) in the equation below, we get;

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

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

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

=312/25(Answer)

 

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

Sol.

Given here; cosθ=3/5 …. (1)

We need to find the value of :\(\frac{\sin \Theta – \frac{1}{\tan \Theta }}{2\tan \Theta }\)   

Now we know from the trigonometric identity;

cos2θ  + sin2θ = 1

By putting the value of cosθ from equation (1), we get;

(3/5)2 + sin2θ = 1

Therefore,

sin2θ = 1 – (3/5)2

sin2θ = 1 – 9/25

sin2θ = (25-9)/25

sin2θ = 16/25

Taking square root on both the sides of above equation;

sinθ = 4/5 ……. (2)

As we know that;

tan θ = sinθ/cosθ

Therefore, by Putting the value of sinθ and cosθ from equation (1) and (2), we get;

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

Now, by Putting the value of sinθ and tan θ from equation (2) and (4), in the given expression;

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

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

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

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

Answer.

 

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

Sol.

Given here;

sinθ=3/5 …….. (1)

We need to find the value of :\(\frac{\cos \Theta – \frac{1}{\tan \Theta }}{2 \cot \Theta }\)

Now, we know from the trigonometric identity;

cos2θ  + sin2θ = 1

Now by putting the value of cosθ from equation (1), we get;

cos2θ + (3/5)2 = 1

or

cos2θ = 1 – (3/5)2

cos2θ = 1 – 9/25

By taking L.C.M of R.H.S;

cos2θ =25-9/25 = 16/25

Now by taking square roots on both the sides, wee get;

cosθ = 4/5 ………. (2)

As we know from the trigonometry ratios;

tanθ = sinθ/cosθ

Now put the value of sinθ and cosθ from equation (1) and (2), we will get;

tanθ = 3/4 …. (3)

Also, we know that;

cot θ = 1/tan θ

Therefore from equation (3), we get;

cot θ = 4/3 …. (4)

Now if we putting the value of cosθ, tanθ and cotθ from equation (2) ,(3) and (4) in the given expression, we will get;

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

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

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

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

= -1/5(Answer)

 

21.)   If tanθ=24/7, find sinθ+ cosθ.

Solution:

Given here;

tanθ=24/7 …. (1)

To find: sinθ+ cosθ

Now, from trigonometric ratios, we know that ;

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

On comparing equation (1) and (2), we get;

Perpendicular side opposite to ∠θ = 24

Base side adjacent to ∠θ= 7

Now, we can construct the triangle with ∠θ, as shown below;

13

Side AC(hypotenuse) is unknown and so by using Pythagoras theorem;

Therefore,

AC2= AB2 + BC2

We know the values of AB and BC from the figure. Thus, we can write the above equation as;

AC2= 242 +72

AC2 = 576 + 49

AC= 625

By taking the square root on both sides, we get;

AC = 25

Therefore, Hypotenuse side AC = 25 …. (3)

Now we know sinθ is equal to;

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

From the figure and equation (3) we can write;

 sinθ = AB/AC

 sinθ = 24/25 ……. (4)

Now, cosθ is equal to;

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

Thus, by putting the value of sinθ and cosθ from equation (4) and (5), we get here;

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

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

Therefore,  sinθ+ cosθ = 31/35

 

22.)    If sinθ = a/b, find sec θ+tanθ in terms of a and b.

Solution:

Given here;

sinθ = a/b ……. (1)

To find: sec θ+tanθ

Now we know, sinθ is defined as follows

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

On comparing equation (1) and (2), we get;

Perpendicular side opposite to ∠θ= a

Hypotenuse = b

Now, we can construct the triangle with ∠θ, as shown below;

14

We can see from the figure, BC is unknown to us.

Therefore, from Pythagoras theorem, we know;

AC2 = AB2 +BC2

Now, by putting the value of AB and AC;

b2 = a2 + BC2

Therefore,

BC2 = b2 – a2

Or

BC= \(\sqrt{b^{2} – a^{2}}\) ………. (3)

Now as we know;

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

Therefore from the figure and equation (3), we cna write;

cosθ = BC/AC

or

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

As we know, secθ = 1/cosθ

so we can write;

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

Now we also know that;

tanθ = sinθ/cosθ

On putting the values from equation (1) and (3), we get;

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

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

On putting the values of secθ and tanθ from equation (5) and (6), 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)

or we can write as, using a22b = (a-b)(a+b), as;

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

We can express the numerator here, as;

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

Now, we can cancel \(\sqrt{b + a}\) from numerator and denominator and write as;

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

Hence, this is the final answer.

 

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

Solution:

Given here;

8tan A = 15

We can write it as;

tan A = 15/8 ………. (1)

To find: sin A – cos A

Now, we know tan A is equal to;

tan A = Perpendicular/Base ………… (2)

If we compare equation (1) and (2), we get;

The perpendicular side opposite to ∠A = 15

Base side adjacent to ∠A = 8

Now, we can construct the triangle with ∠A, as shown below;

15

Hypotenuse side AC is unknown to us, so by using Pythagoras theorem, we get;

AC2= AB2 + BC2

Now by Putting the value of known AB and BC;

AC2= 152 + 82

AC2 = 225 +64

AC2 = 289

Or

AC = √289

AC = 17

Thus, Hypotenuse, AC=17 …….…. (3)

Now we know, sin A is equal to;

sin A = Perpendicular/hypotenuse

So, sin A= BC/AC

Putting the values of BC and AC, we get;

sin A= 15/17 ……..…. (4)

Now we also know, cos A is equal to;

cos A = Base/Hypotenuse

cos A = AB/AC

cos A = 8/17 …….…. (5)

Now we have to find the value of sin A – cos A.

Hence, by putting the value of sin A and cos A from equation (4) and (5), we get;

sin A – cos A = 15/17 – 8/17

sin A – cos A = 15 – 8/17

sin A – cos A = 7/17 (Answer)

 

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

Solution:
Given here;

tan θ = 20/21

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

As we know,

tan θ = Perpendicular/Base

Therefore,

tan θ = 20/21

Let us draw a triangle based on this ratio.

16
You can see from the above figure, AC is the hypotenuse which is unknown to us.

Now, by using Pythagoras theorem;

AC2 = AB2 + BC2

AC2= 212 + 202

AC2 = 441 + 400

AC2 = 841

or

AC = √841

AC = 29

Thus, Hypotenuse, AC= 29

We know from the defintion of sine;

sin θ = Perpendicular/Hypotenuse

So we can say, from the figure;

sin θ = AB/AC

sin θ = 20/29

We know from the defintion of cosine;

cos θ = Base/Hypotenuse

So we can say, from the figure;

cos θ = AB/AC

cos θ = 21/29

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

So by putting the value of sinθ and cosθ we obtained in the above expression;

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

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

After evaluation, we get;

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

Answer.

 

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

Solution:.

Given here;

cosec A = 2

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

We know from the defintion of cosecant;

cosec A = Hypotenuse/Opposite side = 2/1

Let us draw a triangle based on this ratio.

17

Side BC, adjacent ot angle A, is unknown to us. Thus, by using Pythagoras theorem;

AC2= AB2 + BC2

4 = 1 + BC2

BC2= 3

BC = √3

As we already know, from trigonometric ratios;

sin A = 1/cosec A

sin A = 1/2 ……… (1)

And,

tan A= AB/BC

tan A= 1/√3 ………. (2)

And,

cos A = BC/AC

cos A = √3/2 ………. (3)

Putting the values of sin A , cos A and tan A from the equations(1) ,(2) and (3), 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 (Answer)

 

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

Solution:

Given here;

∠A and ∠B are acute angles

And cos A = cos B

To prove: ∠A = ∠B

Let us take a right-angled triangle ACB, right-angles at C, as shown below;

18

Now, as it is given already, cos A = cos B

Thus,

AC/AB = BC/AB

Since, L.H.S and R.H.S has the same denominator,

So, AC=BC

We know already, when two sides of the triangle are equal, then the angles opposite to the sides are also equal.

Therefore,

Angle opposite to side AC = angle opposite to side BC

Or

∠B = ∠A

Or

∠A = ∠B

Hence, proved.

 

27.)  In an △ABC , right-angled triangle at A, if tan C = √3, find the value of sin B cos C + cos B sin C.

Solution:

Given here;

△ABC is a right-angled triangle at A

and tan C = √3 = √3/1 = Perpendicular /Base

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

Let us draw the figure of right-angled △ABC.

19

As you can see, from the figure, sides AB and AC is known to us but side BC is unknown. Thus, by using Pythagoras theorem,

BC2= AB2 +AC2

BC2= √32 + 12
BC2 = 3 +1

BC2 = 4

or

BC = √4

BC = 2

Thus, Hypotenuse side BC= 2 ………… (1)

Now, sin B = Perpendicular/Hypotenuse

So,

sin B = AC/BC

Putting the values of AC and BC, we get;

sin B = 1/2 ……..…. (2)

Now, we knwo, cos B = Base/Hypotenuse

So,

cos B = AB/BC

Putting the values of AB and BC, we get;

cos B= √3/2 …….…. (3)

In the same way,

sin C = √3/2 …..…. (4)

Also,

cos C = AC/BC

cos C = 1/2……………… (5)

Putting the value of sinB, cosB,sin C and cosC from equation (2) ,(3) ,(4) and (5), in sinB cosC + cosB sin C, respectively, we get;

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 (Answer)

 

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

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

             (ii)       sec A = 12/5 for some value of ∠A.

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

           (iv) sin θ = 4/3 for some angle θ.

 

Solution:

(i) We have to justify: tan A < 1

We know, the value of tan A at 45o i.e. tan 45 = 1

As the value of A reaches to 90o

value of Tan A becomes infinite

So the statement is false.

 

(ii) sec A = 12/5, for some value of∠A

If;

sec A = 2.4

sec A > 1

So given statements is true.

If;

For sec A =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. Because cos A is used for cosine of angle A, not for cosecant of angle A.

 

(iv) sin θ = 4/3 for some angle θ.

The given statement is false.

Since the value of sin θ can go to less than or equal to one, for any angle, but here the value of sinθ is more than one.

 

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

Sol.

Given here; sin θ=12/13

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

Since, sin θ=12/13 = Perpendicular/Hypotenuse 

Therefore, Perpendicular = 12 and Hypotenuse  = 13.

Now we can draw a right-angled triangle, whose Perpendicular is 12 unit and Hypotenuse is 13 unit, as shown in the figure;

20

Base of the triangle, BC is unknown to us. Thus, by applying Pythagoras theorem, we get;

AC2=AB2+BC2

169 = 144 +BC2

BC2= 169 – 144

BC2= 25

BC = 5

We know;

cos θ = Base adjacent to angle θ/Hypotenuse

or

cos θ = BC/AC

or

cos θ = 5/13

Also,

tan θ = sin θ/cos θ

Putting the value of sinθ and cosθ in above equations;

tan θ = 12/5

Putting the values of sinθ , cosθ and tan θ 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 simplification, 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,

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

 

30.)   If cos θ = 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 here; If cos θ = 5/13 ……..(1)

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

As we know;

cos θ = Base adjacent to angle θ/Hypotenuse ….(2)

If we compare equation (1) and (2), we get;

Base side adjacent to ∠θ= 5

Hypotenuse = 13

Now we can draw a right-angled triangle, whose base is 5unit and Hypotenuse is 13 unit, as shown in the figure;

21

AB is the perpendicular, which is unknown, so by using Pythagoras theorem, we get;

AC2= AB2 + BC2

Putting the values of base and Hypotenuse we get;

AB2 = 132 – 52

AB2= 169 -25

AB2 = 144

AB = 12 …. (3)

From the figure, we can write;

sin θ= 12/13 …. (4)

Since,

tan θ = sin θ/cos θ

tan θ = 12/5………… (5)

Putting the values of sin θ, cos θ and tan θ 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, on further simplification, we get;

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

Hence,

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

 

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

Solution:

Given here; sec A = 17/8

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

We know, cos A = 1/sec A

Putting the value of sec A, we get;

cos A = 8/17

Also,

cos2 A + sin2 A= 1;

Therefore

sin2 A = 1 – cos2 A 

sin2 A= (8/17)2

sin2 A=225/289

or

sin A= 15/17

Since, tan A = sin A/cos A

Putting the values of sin A and cos A ,we get;

tan A =15/8

Now, put the values of sin A and cos A in the below expression;

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

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

= \(\frac{867 – 900}{256 – 867}\)

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

Again, put the values of tan A in the below expression;

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

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,

Hence from L.H.S and R.H.S it is proved 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 θ = 3/4, prove that \(\sqrt{\frac{cosec^{2}\Theta – \cot ^{2}\Theta }{\sec ^{2} – 1}} = \frac{\sqrt{7}}{3}\)

Sol.

Given here; sin θ = 3/4…. (1)

To prove:

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

We know;

sin  θ = Perpendicular/Hypotenuse …. (3)

On comparing (1) and (3) equation;

Perpendicular = 3 and

Hypotenuse = 4

22

Side BC is unknown and by using Pythagoras theorem to right angled △ABC.

Hence,

AC2 = AB2 +BC2

Now, we can put 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 = √7

Hence, Base side BC =√7 …….. (3)

Now, cos θ = BC/AC √7/4 …..…. (4)

Since, cosec θ = 1/sin A

Thus, cosec θ = 4/3 ………. (5)

In the same way;

sec θ = 4/√7 ……..…. (6)

Since,

cot θ = cos θ/sin θ

Putting the values from eq. (1) and (4),

We get,

cot θ= √7/3 ……..…. (7)

Putting the value of cosec θ, sec θ and cot θ from the equations  (5), (6), and (7) in the L.H.S of the given expression, we get;

\(\sqrt{\frac{cosec^{2}\Theta – \cot ^{2}\Theta }{\sec ^{2} – 1}}\) = \(\sqrt{\frac{\left (\frac{4}{3} \right )^{2} -\left ( \frac{\sqrt{7}}{3} \right )^{2}}{\left (\frac{4}{\sqrt{7}} \right )^{2} – 1}}\)

= \(\frac{\frac{16}{9}-\frac{7}{9}}{\frac{16}{7}-1}\)

= \(\frac{\sqrt{7}}{3}\) R.H.S.

Hence, proved.

 

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

Solution:

Given here; sec A = 17/8 …..…. (1)

To prove:

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

We know, sec A = 1/cos A

So, cos A= 1/sec A

Hence, cos A= 8/17 …..…. (3)

In the same way, we can also get,

sin A = 15/17 …….. (4)

And  tan A= sin A/cos A

tan A= 15/8 ………. (5)

Now, taking the L.H.S;

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

Putting the value of cos A and sin A from equation (3) and (4);

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)

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

Putting 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)

From equation (6) and (7), we get;

L.H.S.=R.H.S

Hence, proved.

 

34.) If cot θ = 3/4, prove that \(\frac{\sec \Theta – cosec\Theta }{\sec \Theta + cosec\Theta } = \frac{1}{\sqrt{7}}\)    

Sol.

Given here; cot θ = 3/4

To prove: \(\frac{\sec \Theta – cosec\Theta }{\sec \Theta + cosec\Theta } = \frac{1}{\sqrt{7}}\)

Since,

cot θ = Base/Perpendicular

Base = 3 and Perpendicular = 4

23

Now by using Pythagoras theorem,

AC2= AB2 +BC2

AC2 = 16 +9

AC2= 25

AC = 5

In the same way,

sec θ = AC/BC

sec θ = 5/3

And cosec θ = AC/BC

cosec θ = 5/4

Putting the values in the given equations we get;

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

Hence, proved.

 

35.) If 3cos θ – 4sin θ = 2cos θ + sin θ, find tanθ.      

Solution.

Given here; 3cos θ – 4sin θ = 2cos θ + sin θ

To find: tanθ?

On solving the given equation;

cos θ = 5 sin θ

On dividing both the sides by cosθ, we get;

cos θ/cos θ = 5 sin θ/cos θ

1 = 5tan θ

tan θ = 1/5. Answer.

 

36.) If ∠A and ∠P are acute angles such that tan A = tan P, then show ∠A= ∠P.

Solution:

Given here; A and P are acute angles tan A =tan P

To prove: ∠A = ∠P

Assume a right-angled triangle ACP, right-angled at C.

24

As we know;

tan θ = opposite side/adjacent side

tan A = PC/AC

tan P =AC/PC

Therefore, tan A =tan P

PC/AC=AC/PC

Or

PC =AC

As we know, the angle opposite to equal sides of a triangle are equal]

Hence,

∠A = ∠P

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