# RD Sharma Solutions for Class 10 Maths Chapter 5 Trigonometric Ratios

### RD Sharma Solutions for Class 10 Maths Chapter 5 Exercise 5.1 Page No: 5.23

1. In each of the following, one of the six trigonometric ratios s given. Find the values of the other trigonometric ratios.

(i) sin A = 2/3

Solution:

We have,

sin A = 2/3 ……..….. (1)

As we know, by sin definition;

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

By comparing eq. (1) and (2), we have

Opposite side = 2 and Hypotenuse = 3

Now, on using Pythagoras theorem in Δ ABC

AC= AB2 + BC2

Putting the values of perpendicular side (BC) and hypotenuse (AC) and for the base side as (AB), we get

⇒ 32 = AB2 + 22

AB= 32 – 22

AB2 = 9 – 4

AB2 = 5

AB = √5

Hence, Base = √5

By definition,

cos A = Base/Hypotenuse

⇒ cos A = √5/3

Since, cosec A = 1/sin A = Hypotenuse/Perpendicular

⇒ cosec A = 3/2

And, sec A = Hypotenuse/Base

⇒ sec A = 3/√5

And, tan A = Perpendicular/Base

⇒ tan A =  2/√5

And, cot A = 1/ tan A = Base/Perpendicular

⇒ cot A = √5/2

(ii) cos A = 4/5

Solution:

We have,

cos A = 4/5 …….…. (1)

As we know, by cos defination

cos A = Base/Hypotenuse …. (2)

By comparing eq. (1) and (2), we get

Base = 4 and Hypotenuse = 5

Now, using Pythagoras theorem in Δ ABC

AC= AB+ BC2

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

52 = 4+ BC2

BC2 = 52 – 42

BC= 25 – 16

BC2 = 9

BC= 3

Hence, Perpendicular = 3

By definition,

sin A = Perpendicular/Hypotenuse

⇒ sin A = 3/5

Then, cosec A = 1/sin A

⇒ cosec A= 1/ (3/5) = 5/3 = Hypotenuse/Perependicular

And, sec A = 1/cos A

⇒ sec A =Hypotenuse/Base

sec A = 5/4

And, tan A = Perpendicular/Base

⇒ tan A = 3/4

Next, cot A = 1/tan A = Base/Perpendicular

∴ cot A = 4/3

(iii) tan θ = 11/1

Solution:

We have, tan θ = 11…..…. (1)

By definition,

tan θ = Perpendicular/ Base…. (2)

On Comparing eq. (1) and (2), we get;

Base = 1 and Perpendicular = 5

Now, using Pythagoras theorem in Δ ABC.

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

Hence, hypotenuse = √122

By definition,

sin = Perpendicular/Hypotenuse

⇒ sin θ = 11/√122

And, cosec θ = 1/sin θ

⇒ cosec θ = √122/11

Next, cos θ = Base/ Hypotenuse

⇒ cos θ = 1/√122

And, sec θ = 1/cos θ

⇒ sec θ = √122/1 = √122

And, cot θ  = 1/tan θ

∴ cot θ = 1/11

(iv) sin θ = 11/15

Solution:

We have,  sin θ = 11/15 ………. (1)

By definition,

sin θ = Perpendicular/ Hypotenuse …. (2)

On Comparing eq. (1) and (2), we get;

Perpendicular = 11 and Hypotenuse= 15

Now, using Pythagoras theorem in Δ ABC

AC2 = AB2 + BC2

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

152 = AB2 +112

AB2 = 15– 112

AB= 225 – 121

AB= 104

AB = √104

AB= √ (2×2×2×13)

AB= 2√(2×13)

AB= 2√26

Hence, Base = 2√26

By definition,

cos θ = Base/Hypotenuse

∴ cosθ = 2√26/ 15

And, cosec θ = 1/sin θ

∴ cosec θ = 15/11

And, secθ = Hypotenuse/Base

∴ secθ =15/ 2√26

And,  tan θ = Perpendicular/Base

∴ tanθ =11/ 2√26

And,  cot θ = Base/Perpendicular

∴ cotθ =2√26/ 11

(v) tan α = 5/12

Solution:

We have,  tan α = 5/12 …. (1)

By definition,

tan α = Perpendicular/Base…. (2)

On Comparing eq. (1) and (2), we get

Base = 12 and Perpendicular side = 5

Now, using Pythagoras theorem in Δ ABC

AC2 = AB2 + BC2

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

AC2 = 122 + 52

AC2 = 144 + 25

AC2= 169

AC = 13 [After taking sq root on both sides]

Hence, Hypotenuse = 13

By definition,

sin α  = Perpendicular/Hypotenuse

∴ sin α = 5/13

And, cosec α  = Hypotenuse/Perpendicular

∴ cosec α = 13/5

And,  cos α = Base/Hypotenuse

∴ cos α = 12/13

And,  sec α =1/cos α

∴ sec α = 13/12

And, 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

Solution:

We have, sin θ = √3/2 …………. (1)

By definition,

sin θ = Perpendicular/ Hypotenuse….(2)

On Comparing eq. (1) and (2), we get;

Perpendicular = √3 and Hypotenuse = 2

Now, using Pythagoras theorem in Δ ABC

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

By definition,

cos θ = Base/Hypotenuse

∴ cos θ = 1/2

And, cosec θ = 1/sin θ

Or cosec θ= Hypotenuse/Perpendicualar

∴ cosec θ =2/√3

And,  sec θ = Hypotenuse/Base

∴ sec θ = 2/1

And,  tan θ = Perpendicula/Base

∴ tan θ = √3/1

And,  cot θ = Base/Perpendicular

∴ cot θ = 1/√3

(vii) cos θ = 7/25

Solution:

We have, cos θ = 7/25 ……….. (1)

By definition,

cos θ = Base/Hypotenuse

On Comparing eq. (1) and (2), we get;

Base = 7 and Hypotenuse = 25

Now, using Pythagoras theorem in Δ ABC

AC2= AB2 + BC2

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

252 = 7+BC2

BC2 = 252 – 72

BC2 = 625 – 49

BC2 = 576

BC= √576

BC= 24

Hence, Perpendicular side = 24

By definition,

sin θ = perpendicular/Hypotenuse

∴ sin θ = 24/25

Since, cosec θ = 1/sin θ

Also, cosec θ= Hypotenuse/Perpendicualar

∴ cosec θ = 25/24

Since,  sec θ = 1/cosec θ

Also,  sec θ = Hypotenuse/Base

∴ sec θ = 25/7

Since,  tan θ = Perpendicular/Base

Now,  cot = 1/tan θ

So,  cot θ = Base/Perpendicular

∴ cot θ = 7/24

(viii) tan θ = 8/15

Solution:

We have, tan θ = 8/15 …………. (1)

By definition,

tan θ = Perpendicular/Base …. (2)

On Comparing eq. (1) and (2), we get;

Base = 15 and Perpendicular = 8

Now, using Pythagoras theorem in Δ ABC

AC= 152 + 82

AC= 225 + 64

AC2 = 289

AC = √289

AC = 17

Hence, Hypotenuse = 17

By definition,

Since,  sin θ = perpendicular/Hypotenuse

∴ sin θ = 8/17

Since, cosec θ = 1/sin θ

Also, cosec θ = Hypotenuse/Perpendicualar

∴ cosec θ = 17/8

Since,  cos θ = Base/Hypotenuse

∴ cos θ = 15/17

Since,  sec θ = 1/cos θ

Also,  sec θ = Hypotenuse/Base

∴ sec θ = 17/15

Since, cot θ = 1/tan θ

Also,  cot θ = Base/Perpendicular

∴ cot θ = 15/8

(ix) cot θ = 12/5

Solution:

We have, cot θ = 12/5 …………. (1)

By definition,

cot θ = 1/tan θ

cot θ = Base/Perpendicular ……. (2)

On Comparing eq. (1) and (2), we have

Base = 12 and Perpendicular side = 5

Now, using Pythagoras theorem in Δ ABC

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

By definition,

Since,  sin θ = perpendicular/Hypotenuse

∴ sin θ= 5/13

Since, cosec θ = 1/sin θ

Also, cosec θ= Hypotenuse/Perpendicualar

∴ cosec θ = 13/5

Since,  cos θ = Base/Hypotenuse

∴ cos θ = 12/13

Since,  sec θ = 1/cosθ

Also,  sec θ = Hypotenuse/Base

∴ sec θ = 13/12

Since,  tanθ = 1/cot θ

Also,  tan θ = Perpendicular/Base

∴ tan θ = 5/12

(x)  sec θ = 13/5

Solution:

We have, sec θ = 13/5…….… (1)

By definition,

sec θ = Hypotenuse/Base…………. (2)

On Comparing eq. (1) and (2), we get

Base = 5 and Hypotenuse = 13

Now, using Pythagoras theorem in Δ ABC

AC= AB2 + BC2

And. putting the value of base side (AB) and hypotenuse (AC) to get the perpendicular side (BC)

132 = 52 + BC2

BC2 = 13– 52

BC2=169 – 25

BC2= 144

BC= √144

BC = 12

Hence, Perpendicular = 12

By definition,

Since,  sin θ = perpendicular/Hypotenuse

∴ sin θ= 12/13

Since, cosec θ= 1/ sin θ

Also, cosec θ= Hypotenuse/Perpendicualar

∴ cosec θ = 13/12

Since,  cos θ= 1/sec θ

Also,  cos θ = Base/Hypotenuse

∴ cos θ = 5/13

Since,  tan θ = Perpendicular/Base

∴ tan θ = 12/5

Since,  cot θ = 1/tan θ

Also,  cot θ = Base/Perpendicular

∴ cot θ = 5/12

(xi)  cosec θ = √10

Solution:

We have, cosec θ = √10/1   ……..… (1)

By definition,

cosec θ = Hypotenuse/ Perpendicualar …….….(2)

And, cosecθ = 1/sin θ

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

Perpendicular side = 1 and Hypotenuse = √10

Now, using Pythagoras theorem in Δ ABC

AC= 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

By definition,

Since,  sin θ = Perpendicular/Hypotenuse

∴ sin θ = 1/√10

Since,  cos θ = Base/Hypotenuse

∴ cos θ = 3/√10

Since,  sec θ = 1/cos θ

Also, 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

(xii)  cos θ =12/15

Solution:

We have; cos θ = 12/15 ………. (1)

By definition,

cos θ = Base/Hypotenuse……… (2)

By comparing eq. (1) and (2), we get;

Base =12 and Hypotenuse = 15

Now, using Pythagoras theorem in Δ ABC, we get

AC2 = AB2+ BC2

Putting the 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

By definition,

Since,  sin θ = perpendicular/Hypotenuse

∴ sin θ = 9/15 = 3/5

Since, cosec θ = 1/sin θ

Also, cosec θ = Hypotenuse/Perpendicualar

∴ cosec θ= 15/9 = 5/3

Since,  sec θ = 1/cos θ

Also,  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

Also,  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

Solution:

(i) Given: In △ABC, AB = 24 cm, BC = 7cm and ∠ABC = 90o

To find: sin A, cos A

By using Pythagoras theorem in △ABC we have

AC2 = AB2 + BC2

AC2 = 242 + 72

AC2 = 576 + 49

AC2= 625

AC = √625

AC= 25

Hence, Hypotenuse = 25

By definition,

sin A = Perpendicular side opposite to angle A/ Hypotenuse

sin A = BC/ AC

sin A = 7/ 25

And,

cos A = Base side adjacent to angle A/Hypotenuse

cos A = AB/ AC

cos A = 24/ 25

(ii) Given: In △ABC , AB = 24 cm and BC = 7cm and ∠ABC = 90o

To find: sin C, cos C

By using Pythagoras theorem in △ABC we have

AC2 = AB2 + BC2

AC2 = 242 + 72

AC2 = 576 + 49

AC2= 625

AC = √625

AC= 25

Hence, Hypotenuse = 25

By definition,

sin C = Perpendicular side opposite to angle C/Hypotenuse

sin C = AB/ AC

sin C = 24/ 25

And,

cos C = Base side adjacent to angle C/Hypotenuse

cos A = BC/AC

cos A = 7/25

3. In fig. 5.37, find tan P and cot R. Is tan P = cot R?

Solution:

By using Pythagoras theorem in △PQR, we have

PR2 = PQ2 + QR2

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

13= 122 + QR2

QR2 = 132 – 122

QR2 = 169 – 144

QR= 25

QR = √25 = 5

By definition,

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

tan P = QR/PQ

tan P = 5/12 ………. (1)

And,

cot R= Base/Perpendicular

cot R= QR/PQ

cot R= 5/12 …. (2)

When comparing equation (1) and (2), we can see that R.H.S of both the equation is equal.

Therefore, L.H.S of both equations should 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 that, sin A = 9/41 …………. (1)

Required to find: cos A, tan A

By definition, we know that

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

On Comparing eq. (1) and (2), we get;

Perpendicular side = 9 and Hypotenuse = 41

Let’s construct △ABC as shown below,

And, here the length of base AB is unknown.

Thus, 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

⇒ Base of triangle ABC, AB = 40

We know that,

cos A = Base/ Hypotenuse

cos A =AB/AC

cos A =40/41

And,

tan A = Perpendicular/ Base

tan A = BC/AB

tan A = 9/40

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

Solution

We have, 15cot A = 8

Required to find: sin A and sec A

As, 15 cot A = 8

⇒ cot A = 8/15 …….(1)

And we know,

cot A = 1/tan A

Also by definition,

Cot A = Base side adjacent to ∠A/ Perpendicular side opposite to ∠A …. (2)

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

Base side adjacent to ∠A = 8

Perpendicular side opposite to ∠A = 15

So, by using Pythagoras theorem to △ABC, we have

AC2 = AB2 +BC2

Substituting values for sides from the figure

AC2 = 82 + 152

AC2 = 64 + 225

AC2 = 289

AC = √289

AC = 17

Therefore, hypotenuse =17

By definition,

sin A = Perpendicular/Hypotenuse

⇒ sin A= BC/AC

sin A= 15/17 (using values from the above)

Also,

sec A= 1/ cos A

⇒ secA = Hypotenuse/ Base side adjacent to ∠A

∴ 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:

△PQR is right-angled at Q.

PQ = 4cm

RQ = 3cm

Required to find: sin P, sin R, sec P, sec R

Given △PQR,

By using Pythagoras theorem to △PQR, we get

PR2 = PQ2 +RQ2

Substituting the respective values,

PR2 = 42 +32

PR2 = 16 + 9

PR2 = 25

PR = √25

PR = 5

⇒ Hypotenuse =5

By definition,

sin P = Perpendicular side opposite to angle P/ Hypotenuse

sin P = RQ/ PR

⇒ sin P = 3/5

And,

sin R = Perpendicular side opposite to angle R/ Hypotenuse

sin R = PQ/ PR

⇒ sin R = 4/5

And,

sec P=1/cos P

secP = Hypotenuse/ Base side adjacent to ∠P

sec P = PR/ PQ

⇒ sec P = 5/4

Now,

sec R = 1/cos R

secR = Hypotenuse/ Base side adjacent to ∠R

sec R = PR/ RQ

⇒ sec R = 5/3

7. If cot θ = 7/8, evaluate

(i)   (1+sin θ)(1–sin θ)/ (1+cos θ)(1–cos θ)

(ii)  cotθ

Solution:

(i) Required to evaluate:
, given = cot θ = 7/8

Taking the numerator, we have

(1+sin θ)(1–sin θ) = 1 – sin2 θ [Since, (a+b)(a-b) = a2 – b2]

Similarly,

(1+cos θ)(1–cos θ) = 1 – cos2 θ

We know that,

sin2 θ + cos2 θ = 1

⇒ 1 – cos2 θ = sin2 θ

And,

1 – sin2 θ = cos2 θ

Thus,

(1+sin θ)(1 –sin θ) = 1 – sin2 θ = cos2 θ

(1+cos θ)(1–cos θ) = 1 – cos2 θ = sin2 θ

= cos2 θ/ sin2 θ

= (cos θ/sin θ)2

And, we know that (cos θ/sin θ) = cot θ

(cot θ)2

= (7/8)2

= 49/ 64

(ii) Given,

cot θ = 7/8

So, by squaring on both sides we get

(cot θ)2 = (7/8)2

∴ cot θ2 = 49/64

8. If 3cot A = 4, check whether (1–tan2A)/(1+tan2A) = (cos2A – sin2A) or not.

Solution:

Given,

3cot A = 4

⇒ cot A = 4/3

By definition,

tan A = 1/ Cot A = 1/ (4/3)

⇒ tan A = 3/4

Thus,

Base side adjacent to ∠A = 4

Perpendicular side opposite to ∠A = 3

In ΔABC, Hypotenuse is unknown

Thus, by applying Pythagoras theorem in ΔABC

We get

AC= AB2 + BC2

AC2 = 42 + 32

AC2 = 16 + 9

AC2 = 25

AC = √25

AC = 5

Hence, hypotenuse = 5

Now, we can find that

sin A = opposite side to ∠A/ Hypotenuse = 3/5

And,

cos A = adjacent side to ∠A/ Hypotenuse = 4/5

Taking the LHS,

Thus, LHS = 7/25

Now, taking RHS

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

Solution:

Given,

tan θ = a/b

And, we know by definition that

tan θ = opposite side/ adjacent side

Thus, by comparison

Opposite side = a and adjacent side = b

To find the hypotenuse, we know that by Pythagoras theorem that

Hypotenuse2 = opposite side2 + adjacent side2

⇒ Hypotenuse = √(a2 + b2)

So, by definition

sin θ = opposite side/ Hypotenuse

sin θ = a/ √(a2 + b2)

And,

cos θ = adjacent side/ Hypotenuse

cos θ = b/ √(a2 + b2)

Now,

After substituting for cos θ and sin θ, we have

Hence Proved.

10. If 3 tan θ = 4, find the value of

Solution:

Given, 3 tan θ = 4

⇒ tan θ = 4/3

From, let’s divide the numerator and denominator by cos θ.

We get,

(4 – tan θ) / (2 + tan θ)

⇒ (4 – (4/3)) / (2 + (4/3)) [using the value of tan θ]

⇒ (12 – 4) / (6 + 4) [After taking LCM and cancelling it]

⇒ 8/10 = 4/5

∴ = 4/5

11. If 3 cot θ = 2, find the value of

Solution:

Given, 3 cot θ = 2

⇒ cot θ = 2/3

From, let’s divide the numerator and denominator by sin θ.

We get,

(4 –3 cot θ) / (2 + 6 cot θ)

⇒ (4 – 3(2/3)) / (2 + 6(2/3)) [using the value of tan θ]

⇒ (4 – 2) / (2 + 4) [After taking LCM and simplifying it]

⇒ 2/6 = 1/3

∴ = 1/3

12. If tan θ = a/b, prove that

Solution:

Given, tan θ = a/b

From LHS, let’s divide the numerator and denominator by cos θ.

And we get,

(a tan θ – b) / (a tan θ + b)

⇒ (a(a/b) – b) / (a(a/b) + b) [using the value of tan θ]

⇒ (a2 – b2)/b2 / (a2 + b2)/b2 [After taking LCM and simplifying it]

⇒ (a2 – b2)/ (a2 + b2)

= RHS

– Hence Proved

13. If sec θ = 13/5, show that

Solution:

Given,

sec θ = 13/5

We know that,

sec θ = 1/ cos θ

⇒ cos θ = 1/ sec θ = 1/ (13/5)

∴ cos θ = 5/13 ……. (1)

By definition,

cos θ = adjacent side/ hypotenuse ….. (2)

Comparing (1) and (2), we have

Adjacent side = 5 and hypotenuse = 13

By Pythagoras theorem,

Opposite side = √((hypotenuse) 2 – (adjacent side)2)

= √(132 – 52)

= √(169 – 25)

= √(144)

= 12

Thus, opposite side = 12

By definition,

tan θ = opposite side/ adjacent side

∴ tan θ = 12/ 5

From, let’s divide the numerator and denominator by cos θ.

We get,

(2 tan θ – 3) / (4 tan θ – 9)

⇒ (2(12/5) – 3) / (4(12/5) – 9) [using the value of tan θ]

⇒ (24 – 15) / (48 – 45) [After taking LCM and cancelling it]

⇒ 9/3 = 3

∴ = 3

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

Solution:

Given, cos θ = 12/13…… (1)

By definition we know that,

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

When comparing equation (1) and (2), we get

Base side adjacent to ∠θ = 12 and Hypotenuse = 13

From the figure,

Base side BC = 12

Hypotenuse AC = 13

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

Thus by applying Pythagoras theorem,

AC= AB2 + BC2

13= AB2 + 122

Therefore,

AB= 13– 122

AB= 169 – 144

AB2 = 25

AB = √25

AB = 5 …. (3)

Now, we know that

sin θ = Perpendicular side opposite to ∠θ / Hypotenuse

Thus, sin θ = AB/AC [from figure]

⇒ sin θ = 5/13… (4)

And, tan θ = sin θ / cos θ = (5/13) / (12/13)

⇒ tan θ = 12/13… (5)

Taking L.H.S we have

L.H.S = sin θ (1 – tan θ)

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

We get,

15.

Solution:

Given, cot θ = 1/3……. (1)

By definition we know that,

cot θ = 1/ tan θ

And, since tan θ = perpendicular side opposite to ∠θ / Base side adjacent to ∠θ

⇒ cot θ = Base side adjacent to ∠θ / perpendicular side opposite to ∠θ …… (2)

[Since they are reciprocal to each other]

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

Base side adjacent to ∠θ = 1 and Perpendicular side opposite to ∠θ = √3

Therefore, the triangle formed is,

On substituting the values of known sides as AB = √3 and BC = 1

AC= (√3) + 1

AC2 = 3 + 1

AC= 4

AC = √4

Therefore, AC = 2 …   (3)

Now, by definition

sin θ = Perpendicular side opposite to ∠θ / Hypotenuse = AB / AC

⇒ sin θ = √3/ 2 ……(4)

And, cos θ = Base side adjacent to ∠θ / Hypotenuse = BC / AC

⇒ cos θ = 1/ 2 ….. (5)

Now, taking L.H.S we have

Substituting the values from equation (4) and (5), we have

16.

Solution:

Given, tan θ = 1/ √7 …..(1)

By definition, we know that

tan θ = Perpendicular side opposite to ∠θ / Base side adjacent to ∠θ ……(2)

On comparing equation (1) and (2), we have

Perpendicular side opposite to ∠θ = 1

Base side adjacent to ∠θ = √7

Thus, the triangle representing ∠ θ is,

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

By applying Pythagoras theorem, we have

AC= AB2 + BC2

AC= 12 + (√7)2

AC 2 = 1 + 7

AC2 = 8

AC = √8

⇒ AC = 2√2

By definition,

sin θ = Perpendicular side opposite to ∠θ / Hypotenuse = AB / AC

⇒ sin θ = 1/ 2√2

And, since cosec θ = 1/sin θ

⇒ cosec θ = 2√2 …….. (3)

Now,

cos θ = Base side adjacent to ∠θ / Hypotenuse = BC / AC

⇒ cos θ = √7/ 2√2

And, since sec θ = 1/ sin θ

⇒ sec θ = 2√2/ √7 ……. (4)

Taking the L.H.S of the equation,

Substituting the value of cosec θ and sec θ from equation (3) and (4), we get

17. If sec θ = 5/4, find the value of

Solution:

Given,

sec θ = 5/4

We know that,

sec θ = 1/ cos θ

⇒ cos θ = 1/ (5/4) = 4/5 …… (1)

By definition,

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

On comparing equation (1) and (2), we have

Hypotenuse = 5

Base side adjacent to ∠θ = 4

Thus, the triangle representing ∠ θ is ABC.

Perpendicular side opposite to ∠θ, AB is unknown and it can be found by using Pythagoras theorem

By applying Pythagoras theorem, we have

AC= AB2 + BC2

AB= AC2 + BC2

AB2 = 52 – 42

AB2 = 25 – 16

AB = √9

⇒ AB = 3

By definition,

sin θ = Perpendicular side opposite to ∠θ / Hypotenuse = AB / AC

⇒ sin θ = 3/ 5 …..(3)

Now, tan θ = Perpendicular side opposite to ∠θ / Base side adjacent to ∠θ

⇒ tan θ = 3/ 4 ……(4)

And, since cot θ = 1/ tan θ

⇒ cot θ = 4/ 3 ……(5)

Now,

Substituting the value of sin θ, cos θ, cot θ and tan θ from the equations (1), (3), (4) and (5) we have,

= 12/7

Therefore,

18. If tan θ = 12/13, find the value of

Solution:

Given,

tan θ = 12/13 …….. (1)

We know that by definition,

tan θ = Perpendicular side opposite to ∠θ / Base side adjacent to ∠θ …… (2)

On comparing equation (1) and (2), we have

Perpendicular side opposite to ∠θ = 12

Base side adjacent to ∠θ = 13

Thus, in the triangle representing ∠ θ we have,

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

So by applying Pythagoras theorem, we have

AC= 122 + 132

AC 2 = 144 + 169

AC2 = 313

⇒ AC = √313

By definition,

sin θ = Perpendicular side opposite to ∠θ / Hypotenuse = AB / AC

⇒ sin θ = 12/ √313…..(3)

And, cos θ = Base side adjacent to ∠θ / Hypotenuse = BC / AC

⇒ cos θ = 13/ √313 …..(4)

Now, substituting the value of sin θ and cos θ from equation (3) and (4) respectively in the equation below

Therefore,

### RD Sharma Solutions for Class 10 Maths Chapter 5 Exercise 5.2 Page No: 5.41

Evaluate each of the following:

1. sin 45 sin 30 + cos 45 cos 30

Solution:

2. sin 60 cos 30 + cos 60 sin 30

Solution:

3.  cos 60 cos 45 – sin 60 sin 45

Solution:

4. sin30 + sin45 + sin60 + sin90

Solution:

5.  cos30 + cos2 45 + cos60 + cos90

Solution:

6. tan30 + tan2 45 + tan2 60

Solution:

7. 2sin2 30 − 3cos45 + tan60

Solution:

8. sin2 30 cos245 + 4tan2 30 + (1/2) sin2 90 − 2cos2 90 + (1/24) cos20

Solution:

9. 4(sin4 60 + cos30) − 3(tan60∘ − tan2 45) + 5cos2 45

Solution:

10. (cosec2 45 sec30)(sin2 30 + 4cot2 45 − sec2 60)

Solution:

11.  cosec3 30 cos60 tan3 45 sin2 90 sec2 45 cot30

Solution:

12. cot2 30 − 2cos2 60 − (3/4)sec2 45 – 4sec2 30

Solution:

Using trigonometric values, we have

13. (cos0 + sin45 + sin30)(sin90 − cos45 + cos60)

Solution:

(cos0 + sin45 + sin30)(sin90 − cos45 + cos60)

Using trigonometric values, we have

15. 4/cot2 30 + 1/sin2 60 − cos2 45

Solution:

16. 4(sin30 + cos2 60) − 3(cos2 45 − sin2 90) − sin2 60

Solution:

Using trigonometric values, we have

17.

Solution:

Using trigonometric values, we have

18.

Solution:

Using trigonometric values, we have

19.

Solution:

Using trigonometric values, we have

Find the value of x in each of the following: (20-25)

20. 2sin 3x = √3

Solution:

Given,

2 sin 3x = √3

sin 3x = √3/2

sin 3x = sin 60°

3x = 60°

x = 20°

### RD Sharma Solutions for Class 10 Maths Chapter 5 Exercise 5.3 Page No: 5.52

1. Evalute the following:

(i) sin 20o/ cos 70o

(ii) cos 19o/ sin 71o

(iii) sin 21o/ cos 69o

(iv) tan 10o/ cot 80o

(v) sec 11o/ cosec 79o

Solution:

(i) We have,

sin 20o/ cos 70o = sin (90o – 70o)/ cos 70= cos 70o/ cos70o = 1 [∵ sin (90 – θ) = cos θ]

(ii) We have,

cos 19o/ sin 71= cos (90o – 71o)/ sin 71= sin 71o/ sin 71o = 1 [∵ cos (90 – θ) = sin θ]

(iii) We have,

sin 21o/ cos 69= sin (90o – 69o)/ cos 69o = cos 69o/ cos69o = 1 [∵ sin (90 – θ) = cos θ]

(iv) We have,

tan 10o/ cot 80= tan (90o – 10o) / cot 80o = cot 80o/ cos80o = 1 [∵ tan (90 – θ) = cot θ]

(v) We have,

sec 11o/ cosec 79o = sec (90o – 79o)/ cosec 79o = cosec 79o/ cosec 79o = 1

[∵ sec (90 – θ) = cosec θ]

2. Evaluate the following:

Solution:

We have, [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]

= 12 + 12 = 1 + 1

= 2

(ii) cos 48°- sin 42°

Solution:

We know that, cos (90° − θ) = sin θ.

So,

cos 48° – sin 42° = cos (90° − 42°) – sin 42° = sin 42° – sin 42°= 0

Thus the value of cos 48° – sin 42° is 0.

Solution:

We have, [∵ cot (90 – θ) = tan θ and cos (90 – θ) = sin θ]

= 1 – 1/2(1)

= 1/2

Solution:

We have, [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]

= 1 – 1

= 0

Solution:

We have, [∵ cot (90 – θ) = tan θ and tan (90 – θ) = cot θ]

= tan (90o – 35o)/ cot 55o + cot (90o – 12o)/ tan 12o – 1

= cot 55o/ cot 55o + tan 12o/ tan 12o – 1

= 1 + 1 – 1

= 1

Solution:

We have , [∵ sin (90 – θ) = cos θ and sec (90 – θ) = cosec θ]

= sec (90o – 20o)/ cosec 20o + sin (90o – 31o)/ cos 31o

= cosec 20o/ cosec 20o + cos 12o/ cos 12o

= 1 + 1

= 2

(vii) cosec 31° – sec 59°

Solution:

We have,

cosec 31° – sec 59°

Since, cosec (90 – θ) = cos θ

So,

cosec 31° – sec 59° = cosec (90° – 59o) – sec 59° = sec 59° – sec 59° = 0

Thus,

cosec 31° – sec 59° = 0

(viii) (sin 72° + cos 18°) (sin 72° – cos 18°)

Solution:

We know that,

sin (90 – θ) = cos θ

So, the given can be expressed as

(sin 72° + cos 18°) (sin (90 – 18)° – cos 18°)

= (sin 72° + cos 18°) (cos 18° – cos 18°)

= (sin 72° + cos 18°) x 0

= 0

(ix) sin 35° sin 55° – cos 35° cos 55°

Solution:

We know that,

sin (90 – θ) = cos θ

So, the given can be expressed as

sin (90 – 55)° sin (90 – 35)° – cos 35° cos 55°

= cos 55° cos 35° – cos 35° cos 55°

= 0

(x) tan 48° tan 23° tan 42° tan 67°

Solution:

We know that,

tan (90 – θ) = cot θ

So, the given can be expressed as

tan (90 – 42)° tan (90 – 67)° tan 42° tan 67°

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°)(cot 67° tan 67°)

= 1 x 1 [∵ tan θ x cot θ = 1]

= 1

(xi) sec 50° sin 40° + cos 40° cosec 50°

Solution:

We know that,

sin (90 – θ) = cos θ and cos (90 – θ) = sin θ

So, the given can be expressed as

sec 50° sin (90 – 50)° + cos (90 – 50)° cosec 50°

= sec 50° cos 50° + sin 50° cosec 50°

= 1 + 1 [∵ sin θ x cosec θ = 1 and cos θ x sec θ = 1]

= 2

3. Express each one of the following in terms of trigonometric ratios of angles lying between 0o and 45o

(i) sin 59o + cos 56o (ii) tan 65o + cot 49o (iii) sec 76o + cosec 52o

(iv) cos 78o + sec 78o (v) cosec 54o + sin 72o (vi) cot 85o + cos 75o

(vii) sin 67o + cos 75o

Solution:

Using the below trigonometric ratios of complementary angles, we find the required

sin (90 – θ) = cos θ cosec (90 – θ) = sec θ

cos (90 – θ) = sin θ sec (90 – θ) = cosec θ

tan (90 – θ) = cot θ cot (90 – θ) = tan θ

(i) sin 59o + cos 56o = sin (90 – 31)o + cos (90 – 34)o = cos 31o + sin 34o

(ii) tan 65o + cot 49o = tan (90 – 25)o + cot (90 -31)o = cot 25o + tan 31o

(iii) sec 76o + cosec 52o = sec (90 – 14)o + cosec (90 – 38)o = cosec 14o + sec 38o

(iv) cos 78o + sec 78= cos (90 – 12)o + sec (90 – 12)o = sin 12o + cosec 12o

(v) cosec 54o + sin 72o = cosec (90 – 36)o + sin (90 – 18)o = sec 36o + cos 18o

(vi) cot 85o + cos 75o = cot (90 – 5)o + cos (90 – 15)o = tan 5o + sin 15o

4. Express cos 75o + cot 75o in terms of angles between 0o and 30o.

Solution:

Given,

cos 75o + cot 75o

Since, cos (90 – θ) = sin θ and cot (90 – θ) = tan θ

cos 75o + cot 75= cos (90 – 15)o + cot (90 – 15)o = sin 15+ tan 15o

Hence, cos 75o + cot 75o can be expressed as sin 15+ tan 15o

5. If sin 3A = cos (A – 26o), where 3A is an acute angle, find the value of A.

Solution:

Given,

sin 3A = cos (A – 26o)

Using cos (90 – θ) = sin θ, we have

sin 3A = sin (90o – (A – 26o))

Now, comparing both L.H.S and R.H.S

3A = 90o – (A – 26o)

3A + (A – 26o) = 90o

4A – 26o = 90o

4A = 116o

A = 116o/4

∴ A = 29o

6. If A, B, C are the interior angles of a triangle ABC, prove that

(i) tan ((C + A)/ 2) = cot (B/2) (ii) sin ((B + C)/ 2) = cos (A/2)

Solution:

We know that, in triangle ABC the sum of the angles i.e A + B + C = 180o

So, C + A = 180o – B ⇒ (C + A)/2 = 90o – B/2 …… (i)

And, B + C = 180– A ⇒ (B + C)/2 = 90o – A/2 ……. (ii)

(i) L.H.S = tan ((C + A)/ 2)

⇒ tan ((C + A)/ 2) = tan (90o – B/2) [From (i)]

= cot (B/2) [∵ tan (90 – θ) = cot θ]

= R.H.S

• Hence Proved

(ii) L.H.S = sin ((B + C)/2)

⇒ tan ((B + C)/ 2) = tan (90o – A/2) [From (ii)]

= cot (B/2) [∵ tan (90 – θ) = cot θ]

= R.H.S

• Hence Proved

7. Prove that:

(i) tan 20° tan 35° tan 45° tan 55° tan 70° = 1

(ii) sin 48° sec 48° + cos 48° cosec 42° = 2

Solution:

(i) Taking L.H.S = tan 20° tan 35° tan 45° tan 55° tan 70°

= tan (90° − 70°) tan (90° − 55°) tan 45°tan 55° tan70°

= cot 70°cot 55° tan 45° tan 55° tan 70°  [∵ tan (90 – θ) = cot θ]

= (tan 70°cot 70°)(tan 55°cot 55°) tan 45°  [∵ tan θ x cot θ = 1]

= 1 × 1 × 1 = 1

• Hence proved

(ii) Taking L.H.S = sin 48° sec 48° + cos 48° cosec 42°

= sin 48° sec (90° − 48°) cos 48° cosec (90° − 48°)

[∵sec (90 – θ) = cosec θ and cosec (90 – θ) = sec θ]

= sin 48°cosec 48° + cos 48°sec 48°  [∵ cosec θ x sin θ = 1 and cos θ x sec θ = 1]

= 1 + 1 = 2

• Hence proved

(iii) Taking the L.H.S,

= 1 + 1 – 2

= 2 – 2

= 0

• Hence proved

(iv) Taking L.H.S,

= 1 + 1

= 2

• Hence proved

8. Prove the following:

(i) sinθ sin (90o – θ) – cos θ cos (90o – θ) = 0

Solution:

Taking the L.H.S,

sinθ sin (90o – θ) – cos θ cos (90o – θ)

= sin θ cos θ – cos θ sin θ [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]

= 0

(ii)

Solution:

Taking the L.H.S,

[∵ cosec θ x sin θ = 1 and cos θ x sec θ = 1]

= 1 + 1

= 2 = R.H.S

• Hence Proved

(iii)

Solution:

Taking the L.H.S, [∵ tan (90o – θ) = cot θ]

= 0 = R.H.S

• Hence Proved

(iv)

Solution:

Taking L.H.S, [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]

= sin2 A = R.H.S

• Hence Proved

(v) sin (50o + θ) – cos (40o – θ) + tan 1o tan 10o tan 70o tan 80o tan 89o = 1

Solution:

Taking the L.H.S,

= sin (50o + θ) – cos (40o – θ) + tan 1o tan 10o tan 20o tan 70o tan 80o tan 89o

= sin (50o + θ) – sin (90o – (40o – θ)) + tan (90 – 89)o tan (90 – 80)o tan (90 – 70)o tan 70o tan 80o tan 89[∵ sin (90 – θ) = cos θ]

= sin (50o + θ) – sin (50o + θ) + cot 89o cot 80o cot 70o tan 70o tan 80o tan 89o

[∵ tan (90o – θ) = cot θ]

= 0 + (cot 89o x tan 89o) (cot 80o x tan 80o) (cot 70o x tan 70o)

= 0 + 1 x 1 x 1 [∵ tan θ x cot θ = 1]

= 1= R.H.S

• Hence Proved
Courtesy : CBSE