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

The applications of trigonometry are numerous 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 Maths Chapter 5 Trigonometric Ratios Exercise 5.3

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

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

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°)

= 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

= 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