RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3 (Updated for 2024)

RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3: The following RD Sharma Solutions Class 10 exercise 5.3 contains a number of problems to help students better understand this concept. The solutions are written in simple Language to meet the needs of all students. For a full explanation of the questions in this exercise, students can download RD Sharma Solutions for Class 10 Maths Chapter 5 Trigonometric Ratios Exercise 5.3 PDF, which is provided below.

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RD Sharma Class 10 Solutions Chapter 5 Exercise 5.3

Access answers to RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3- Important Question with Answers

1. Evaluate the following:

(i) sin 20^o/ cos 70^o

(ii) cos 19^o/ sin 71^o

(iii) sin 21^o/ cos 69^o

(iv) tan 10^o/ cot 80^o

(v) sec 11^o/ cosec 79^o

Solution:

(i) We have,

sin 20^o/ cos 70^o = sin (90^o – 70^o)/ cos 70^o = cos 70^o/ cos70^o = 1 [∵ sin (90 – θ) = cos θ]

(ii) We have,

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

(iii) We have,

sin 21^o/ cos 69^o = sin (90^o – 69^o)/ cos 69^o = cos 69^o/ cos69^o = 1 [∵ sin (90 – θ) = cos θ]

(iv) We have,

tan 10^o/ cot 80^o = tan (90^o – 10^o) / cot 80^o = cot 80^o/ cos80^o = 1 [∵ tan (90 – θ) = cot θ]

(v) We have,

sec 11^o/ cosec 79^o = sec (90^o – 79^o)/ cosec 79^o = cosec 79^o/ cosec 79^o = 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 (90^o – 35^o)/ cot 55^o + cot (90^o – 12^o)/ tan 12^o – 1

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

= 1 + 1 – 1

= 1

Solution:

 

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

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

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

= 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° – 59^o) – 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 0^o and 45^o

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

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

(vii) sin 67^o + cos 75^o

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

(ii) tan 65^o + cot 49^o = tan (90 – 25)^o + cot (90 -41)^o = cot 25^o + tan 41^o

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

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

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

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

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

Solution:

Given,

cos 75^o + cot 75^o

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

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

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

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

Solution:

Given,

sin 3A = cos (A – 26^o)

Using cos (90 – θ) = sin θ, we have

sin 3A = sin (90^o – (A – 26^o))

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

3A = 90^o – (A – 26^o)

3A + (A – 26^o) = 90^o

4A – 26^o = 90^o

4A = 116^o

A = 116^o/4

∴ A = 29^o

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 = 180^o

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

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

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

⇒ tan ((C + A)/ 2) = tan (90^o – 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)

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

= cos (A/2)

= 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 (90^o – θ) – cos θ cos (90^o – θ) = 0

Solution:

Taking the L.H.S,

sinθ sin (90^o – θ) – cos θ cos (90^o – 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 (90^o – 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 (50^o + θ) – cos (40^o – θ) + tan 1^o tan 10^o tan 20^o tan 70^o tan 80^o tan 89^o = 1

Solution:

Taking the L.H.S,

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

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

= cos (40^o – θ) – cos (40^o – θ) + cot 89^o cot 80^o cot 70^o tan 70^o tan 80^o tan 89^o

[∵ tan (90^o – θ) = cot θ]

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

= 0 + 1 x 1 x 1 [∵ tan θ x cot θ = 1]

= 1= R.H.S

• Hence Proved

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