# Class 10 RD Sharma Solutions – Chapter 5 Trigonometric Ratios – Exercise 5.3 | Set 2

Last Updated : 03 Mar, 2021

### (i) sin Î¸ sin (90Â° – Î¸) – cosÎ¸ cos (90Â° – Î¸) = 0

Solution:

We have to prove that  sin Î¸ sin (90Â° – Î¸) – cosÎ¸ cos (90Â° – Î¸) = 0

Taking LHS

= sin Î¸ sin (90Â° – Î¸) – cosÎ¸ cos (90Â° – Î¸)           -(âˆµ sin (90Â° – Î¸) = cos Î¸)

= sin Î¸ cosÎ¸ – cosÎ¸ sinÎ¸

= 0

LHS = RHS

Hence proved

### (ii)

Solution:

We have to prove that

Taking LHS

= tan Î¸/tan Î¸ + cot Î¸/cot Î¸

= 1 + 1

= 2

LHS = RHS

Hence Proved

### (iii)

Solution:

We have to prove that

Taking LHS

-(âˆµ cot Î¸ = cosÎ¸/sinÎ¸ and cosecÎ¸ = 1/sinÎ¸)

= cos2A – cos2A

= 0

LHS = RHS

Hence Proved

### (iv)

Solution:

We have to prove that

Taking LHS

= sin2A

LHS = RHS

Hence Proved

### (v) sin (50Â° + Î¸) – cos (40Â° – Î¸) + tan 1Â° tan 10Â° tan 20Â° tan 70Â° tan 80Â° tan 89Â° = 1

Solution:

We have to prove that sin (50Â° + Î¸) – cos (40Â° – Î¸) + tan 1Â° tan 10Â° tan 20Â° tan 70Â° tan 80Â° tan 89Â° = 1

Taking LHS

= sin (50Â° + Î¸) – cos (40Â° – Î¸) + tan 1Â° tan 10Â° tan 20Â° tan 70Â° tan 80Â° tan 89Â°

= cos(90Â° – (50Â° + Î¸)) – cos (40Â° – Î¸) + tan (90Â° – 89Â°) tan (90Â° – 80Â°) tan (90Â° – 70Â°) tan 70Â° tan 80Â° tan 89Â°

= cos (40Â° – Î¸) – cos (40Â° – Î¸) + cot 89Â° cot 80Â° cot 70Â° tan 70Â° tan 80Â° tan 89Â°

= 0 + 1 = 1

LHS = RHS

Hence Proved

### (i) 2/3(cos430Â° – sin445Â°) – 3(sin260Â° – sec245Â°) + 1/4cot230Â°

Solution:

Given: 2/3(cos430Â° – sin445Â°) – 3(sin260Â° – sec245Â°) + 1/4cot230Â°

= 2/3(5/16) – 3(-5/4) + 3/4

= 5/24 + 90/24 + 18/24

= 113/24

Hence, 2/3(cos430Â° – sin445Â°) – 3(sin260Â° – sec245Â°) + 1/4cot230Â° = 113/24

### (ii) 4(sin430Â° + cos460Â°) – 2/3(sin260Â° – cos245Â°) + 1/2tan260Â°

Solution:

Given: 4(sin430Â° + cos460Â°) – 2/3(sin260Â° – cos245Â°) + 1/2tan260Â°

= 4(2/16) – 2/3(1/4) + 3/2

= 1/2 – 1/6 + 3/2

= 4/2 – 1/6

= 2 – 1/6

= (12 – 1)/6

= 11/6

Hence, 4(sin430Â° + cos460Â°) – 2/3(sin260Â° – cos245Â°) + 1/2tan260Â° = 11/6

### (iii) sin 50Â°/cos 40Â° + cosec 40Â° / sec 50Â° – 4cos 50Â° cosec40Â°

Solution:

Given: sin 50Â°/cos 40Â° + cosec 40Â° / sec 50Â° – 4cos 50Â° cosec40Â°

= sin 50Â°/sin 50Â° + cosec 40Â° / cosec 40Â° – 4cos 50Â° sec50Â°

= 1 + 1 – 4

= -2

Hence, sin 50Â°/cos 40Â° + cosec 40Â° / sec 50Â° – 4cos 50Â° cosec40Â° = -2

### (iv) tan35Â°tan40Â°tan45Â°tan50Â°tan55Â°

Solution:

Given: tan35Â°tan40Â°tan45Â°tan50Â°tan55Â°

= tan(90Â° – 55Â°)tan(90Â° – 50Â°)tan45Â°tan50Â°tan55Â°

= cot55Â°cot50Â°tan45Â°tan50Â°tan55Â°

= (1/tan55Â°)(1/tan50Â°)tan45Â°tan50Â°tan55Â°

= 1

Hence, tan35Â°tan40Â°tan45Â°tan50Â°tan55Â° = 1

### (v) cosec(65Â° + Î¸) – sec(25Â° – Î¸) – tan(55Â° – Î¸) + cot(35Â° + Î¸)

Solution:

Given: cosec(65Â° + Î¸) – sec(25Â° – Î¸) – tan(55Â° – Î¸) + cot(35Â° + Î¸)

= cosec(65Â° + Î¸) – cosec(90Â° – (25Â° – Î¸)) – tan(55Â° – Î¸) + tan(90Â° – (35Â° + Î¸))

= cosec(65Â° + Î¸) – cosec(90Â° – 25Â° + Î¸) – tan(55Â° – Î¸) + tan(90Â° – 35Â° – Î¸)

= cosec(65Â° + Î¸) – cosec(65Â° + Î¸) – tan(55Â° – Î¸) + tan(55Â° – Î¸)

= 0

Hence, cosec(65Â° + Î¸) – sec(25Â° – Î¸) – tan(55Â° – Î¸) + cot(35Â° + Î¸) = 0

### (vi) tan7Â°tan23Â°tan60Â°tan67Â°tan83Â°

Solution:

Given: tan7Â°tan23Â°tan60Â°tan67Â°tan83Â°

= tan(90Â° – 83Â°)tan(90Â° – 67Â°)tan60Â°tan67Â°tan83Â°

= cot83Â°cot67Â°tan60Â°tan67Â°tan83Â°

= (1/tan83Â°)(1/tan67Â°)tan60Â°tan67Â°tan83Â°

= tan60Â°

= âˆš3

Hence, tan7Â°tan23Â°tan60Â°tan67Â°tan83Â° = âˆš3

### (vii)

Solution:

Given:

= 2 – 2/5 – 3/5

= (10 – 2 – 3)/5

= 5/5

= 1

Hence,  = 1

Solution:

Given:

= 3/7 – 4/7

= -1/7

Hence,  = -1/7

### (ix) sin18Â°/cos72Â° + âˆš3(tan10Â°tan30Â°tan40Â°tan50Â°tan80Â°)

Solution:

Given: sin18Â°/cos72Â° + âˆš3(tan10Â°tan30Â°tan40Â°tan50Â°tan80Â°)

= sin18Â°/cos(90Â° – 18Â°) + âˆš3(tan(90Â° – 80Â°)tan(90Â° – 50Â°)tan30Â°tan50Â°tan80Â°)

= sin18Â°/sin18Â° + âˆš3(cot80Â°cot50Â°tan30Â°tan50Â°tan80Â°)

= 1 + âˆš3tan30Â°

= 1 + âˆš3(1/âˆš3)

= 2

Hence, sin18Â°/cos72Â° + âˆš3(tan10Â°tan30Â°tan40Â°tan50Â°tan80Â°) = 2

### (x)

Solution:

Given:

= 1 + 1 – 1/tan60Â°

= 2 – 1/âˆš3

= (2âˆš3 – 1)/âˆš3

= (6 – âˆš3)/3

Hence, = (6 – âˆš3)/3

### Question 10. If sin Î¸ = cos(Î¸ âˆ’ 45Â°), where Î¸ and Î¸âˆ’45Â° are acute angles, find the degree measure of Î¸.

Solution:

Given: sinÎ¸ = cos(Î¸ âˆ’ 45Â°), where Î¸ and Î¸âˆ’45Â° are acute angles

= sinÎ¸ = cos(Î¸ âˆ’ 45Â°)           -(âˆµ sinÎ¸ = cos(90 âˆ’ Î¸))

= cos(90Â° âˆ’ Î¸) = cos(Î¸ âˆ’ 45Â°)

Equating the angles

(90Â° âˆ’ Î¸) = (Î¸ âˆ’ 45Â°)

2Î¸ = 90Â° + 45Â° = 135Â°

Î¸ = 135Â°/2

Î¸ = 67Â°

### (i) sin (B + C)/2 = cos A/2

Solution:

According to question

In a triangle ABC, A, B, C are the interior angles

So, the sum of interior angles = A + B + C = 180Â°

B +C  = 180Â° – A

We have

sin (B + C)/2 = cos A/2

Taking LHS

sin (B + C)/2          -(1)

Putting the value of B + C in equation(1)

= sin (180Â° – A)/2

= sin (90Â° – A/2)

= cos A/2

LHS = RHS

Hence Proved

### (ii) cos (B + C)/2 = sin A/2

Solution:

According to question

In a triangle ABC, A, B, C are the interior angles

So, the sum of interior angles = A + B + C = 180Â°

B + C = 180Â° – A

We have

cos (B + C)/2 = sin A/2

Taking LHS

cos (B + C)/2          -(1)

Putting the value of B + C in equation(1)

= cos (180Â° – A)/2

= cos (90Â° – A/2)

= sin A/2

LHS = RHS

Hence Proved

### Question 12. If 2Î¸ + 45Â° and 30Â° âˆ’ Î¸ are acute angles, find the degree measure of Î¸ satisfying sin(2Î¸ + 45Â°) = cos(30Â° âˆ’ Î¸).

Solution:

Given: 2Î¸ + 45Â° and (30Â° âˆ’ Î¸) are acute and sin(2Î¸ + 45Â°) = cos(30Â° âˆ’ Î¸)

We have,

sin(2Î¸ + 45Â°) = cos(30Â° âˆ’ Î¸)

sin(2Î¸ + 45Â°) = sin(90Â° âˆ’ (30Â° âˆ’ Î¸))           -(âˆµ cosÎ¸ = sin(90Â° âˆ’ Î¸))

sin(2Î¸ + 45Â°) = sin(60Â° + Î¸)

Now equating the angles

2Î¸ + 45Â° = 60Â° + Î¸

2Î¸ âˆ’ Î¸ = 60Â° âˆ’ 45Â°

Î¸ = 15Â°

### Question 13. If Î¸ is a positive acute angle such that secÎ¸ = cosec60Â°, find the value of 2cos2Î¸ âˆ’ 1.

Solution:

Given, Î¸ is acute and secÎ¸ = cosec60Â°

Find the value of 2cos2Î¸ âˆ’ 1       -(1)

cosec60Â° = 2/âˆš3

or secÎ¸ = 2/âˆš3

We know that, sec30Â° = 2/âˆš3

secÎ¸ = sec30Â°

Î¸ = 30Â°

Putting the value of Î¸ in eq(1), we get

= 2cos230Â° âˆ’ 1

= 2(âˆš3/2)2 âˆ’ 1

= 2(3/4) âˆ’ 1

= 3/2 âˆ’ 1

= 1/2

Hence, the value of 2cos2Î¸ âˆ’ 1 = 1/2

### Question 14. If cos2Î¸ = sin4Î¸, where 2Î¸ and 4Î¸ are acute angles, find the value of Î¸.

Solution:

Given: 2Î¸ and 4Î¸ are acute and cos2Î¸ = sin4Î¸

So, we have,

cos2Î¸ = sin(90Â° âˆ’ 2Î¸)           -(âˆµ sin (90Â° – Î¸) = cos Î¸)

Now, sin(90Â° âˆ’ 2Î¸) = sin4Î¸

Equating the angles

90Â° âˆ’ 2Î¸ = 4Î¸

90Â° = 2Î¸ + 4Î¸

6Î¸ = 90Â°

Î¸ = 15Â°

Hence, the value of Î¸ = 15Â°

### Question 15. If sin3Î¸ = cos (Î¸ âˆ’ 6Â°), where 3 Î¸ and Î¸ âˆ’ 6Â° are acute, find the value of Î¸.

Solution:

Given: 3Î¸ and (Î¸ âˆ’ 6Â°) are acute and sin3Î¸ = cos(Î¸ âˆ’ 6Â°)

So, we have,

cos(Î¸ âˆ’ 6Â°) = sin(90Â° âˆ’ (Î¸ âˆ’ 6Â°)) = sin(96Â° âˆ’ Î¸).         -(âˆµ cosÎ¸ = sin(90Â° âˆ’ Î¸))

Now, sin3Î¸ = sin(96Â° âˆ’ Î¸)

Equating the angles

3Î¸ = 96Â° âˆ’ Î¸

3Î¸ + Î¸ = 96Â°

4Î¸ = 96Â°

Î¸ = 96Â°/4

Î¸ = 24Â°

Hence, the value of Î¸ = 24Â°

### Question 16. If sec4A = cosec(A – 20Â°), where 4A is an acute angle, find the value of A.

Solution:

Given: 4A is acute and sec4A = cosec(A âˆ’ 20Â°)

So, we have,

sec4A = cosec(90Â° âˆ’ 4A)         -(âˆµ secÎ¸ = cosec(90Â° âˆ’ Î¸))

Now, cosec(90Â° âˆ’ 4A) = cosec(A âˆ’ 20Â°)

Equating the angles

(90Â° âˆ’ 4A) = (A âˆ’ 20Â°)

110Â° = 5A

5A = 110Â°

A = 110Â°/5

A = 22Â°

Hence, the value of A = 22Â°

### Question 17. If sec2A = cosec(A – 42Â°), where 2A is an acute angle, find the value of A.

Solution:

Given: 2A is acute and sec2A = cosec(A âˆ’ 42Â°)

So, we have,

sec2A = cosec(90Â° âˆ’ 2A)          -(âˆµ secÎ¸ = cosec(90Â° âˆ’ Î¸))

Now, cosec(90Â° âˆ’ 2A) = cosec(A âˆ’ 42Â°)

Equating the angles

(90Â° âˆ’ 2A) = (A âˆ’ 42Â°)

132Â° = 3A

3A = 132Â°

A = 132Â°/3

A = 44Â°

Hence, the value of A = 44Â°

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