RD Sharma Solutions for Class 10 Chapter 5 Trigonometric Ratios Exercise 5.3

RD Sharma Class 10 Solutions Chapter 5 Ex 5.3 PDF Free Download

The applications of trigonometry are enormous in the real world. One of the interesting concepts in trigonometry is the trigonometric ratios of complementary angles. The following RD Sharma Solutions Class 10 exercise 5.3 has several problems to help strengthen this concept for the students. Solutions prepared are in simple language to suit the needs of every student. The students can also download RD Sharma Solutions for Class 10 Maths Chapter 5 Trigonometric Ratios Exercise 5.3 PDF provided below for a detailed explanation of problems in this exercise.

RD Sharma Solutions for Class 10 Chapter 5 Trigonometric Ratios Exercise 5.3 Download PDF

 

rd sharma solutions for class 10 chapter 5 ex 3
rd sharma solutions for class 10 chapter 5 ex 3
rd sharma solutions for class 10 chapter 5 ex 3
rd sharma solutions for class 10 chapter 5 ex 3
rd sharma solutions for class 10 chapter 5 ex 3
rd sharma solutions for class 10 chapter 5 ex 3
rd sharma solutions for class 10 chapter 5 ex 3
rd sharma solutions for class 10 chapter 5 ex 3
rd sharma solutions for class 10 chapter 5 ex 3

 

Access RD Sharma Solutions for Class 10 Chapter 5 Trigonometric Ratios Exercise 5.3

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

(ii) We have,

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

(iii) We have,

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

(iv) We have,

tan 10o/ cot 80o = 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:

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 1

Solution:

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

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 2

= 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.

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 3

Solution:

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

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 4

= 1 – 1/2(1)

= 1/2

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 5

Solution:

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

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 6

= 1 – 1

= 0

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 6

Solution:

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

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 8

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

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

= 1 + 1 – 1

= 1

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 9

Solution:

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

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 10

= 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 78o = 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 75o = cos (90 – 15)o + cot (90 – 15)o = sin 15o + tan 15o

Hence, cos 75o + cot 75o can be expressed as sin 15o + 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 = 180o – 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

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 11

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,

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 12

= 1 + 1 – 2

= 2 – 2

= 0

  • Hence proved

(iv) Taking L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 13

= 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

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 14(ii)

Solution:

Taking the L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 15

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 16[∵ cosec θ x sin θ = 1 and cos θ x sec θ = 1]

= 1 + 1

= 2 = R.H.S

  • Hence Proved

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 17(iii)

Solution:

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

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 18

= 0 = R.H.S

  • Hence Proved

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 19(iv)

Solution:

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

R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.3 - 20

= sin2 A = R.H.S

  • Hence Proved

(v) sin (50o + θ) – cos (40oθ) + 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 89o [∵ 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

Also, access other exercise solutions of RD Sharma Class 10 Maths Chapter 5 Trigonometric Ratios

Exercise 5.1 Solutions

Exercise 5.2 Solutions

Leave a Comment

Your email address will not be published. Required fields are marked *