# Class 10 RD Sharma Solutions – Chapter 5 Trigonometric Ratios – Exercise 5.3 | Set 1

Last Updated : 03 Mar, 2021

### (i) sin 20Â°/cos 70Â°

Solution:

Given: sin 20Â°/cos 70Â°

= sin(90Â° âˆ’ 70Â°)/cos 70Â°

= cos 70Â°/cos 70Â°          -(âˆµ sin (90Â° – Î¸) = cos Î¸)

= 1

Hence, sin 20Â°/cos 70Â° = 1

### (ii) cos 19Â°/sin 71Â°

Solution:

Given: cos 19Â°/sin 71Â°

= cos(90Â° âˆ’ 71Â°)/sin 71Â°

= sin 71Â°/sin 71Â°             -(âˆµ cos (90Â° – Î¸) = sin Î¸)

= 1

Hence, cos 19Â°/sin 71Â° = 1

### (iii) sin 21Â°/cos 69Â°

Solution:

Given: sin 21Â°/cos 69Â°

= sin(90Â° âˆ’ 69Â°)/cos 69Â°

= cos 69Â°/cos 69Â°            -(âˆµ sin (90Â° – Î¸) = cos Î¸)

= 1

Hence, sin 21Â°/cos 69Â° = 1

### (iv) tan 10Â°/cot 80Â°

Solution:

Given: tan 10Â°/cot 80Â°

= tan(90Â° âˆ’ 80Â°)/cot 80Â°

= cot 80Â°/cot 80Â°            -(âˆµ tan (90Â° – Î¸) = cot Î¸)

= 1

Hence, tan 10Â°/cot 80Â° = 1

### (v) sec 11Â°/cosec 79Â°

Solution:

Given: sec 11Â°/cosec 79Â°

= sec(90Â° âˆ’ 79Â°)/cosec 79Â°

= cosec 79Â°/cosec 79Â°          -(âˆµ sec (90Â° – Î¸) = cosec Î¸)

= 1

Hence, sec 11Â°/cosec 79Â° = 1

### (i) ()2 + ()2

Solution:

Given: ()2 + ()

-(âˆµ sin (90Â° – Î¸) = cos Î¸)

= (cos 41Â°/cos 41Â°)2 + (cos 41Â°/cos 41Â°)2

= 1 + 1 = 2

Hence, ()2 + ()2 = 2

### (ii) cos 48Â° – sin 42Â°

Solution:

Given: cos 48Â° – sin 42Â°

= cos 48Â° – sin (90Â°- 48Â°)           -(âˆµ sin (90Â° – Î¸) = cos Î¸)

= cos 48Â° – cos 48Â°

= 0

Hence, cos 48Â° – sin 42Â° = 0

### (iii)

Solution:

Given:

-(âˆµ sin (90Â° – Î¸) = cos Î¸ and cot (90Â° – Î¸) = tan Î¸)

= 1 – 1/2 = 1/2

Hence,  = 1/2

### (iv) ()2– ()2

Solution:

Given: ()2– ()2

-(âˆµ sin (90Â° – Î¸) = cos Î¸)

= (cos 63Â°/cos 63Â°)2 – (cos 63Â°/cos 63Â°)2

= 1 – 1 = 0

Hence, ()2– ()2 = 0

### (v)

Solution:

Given:

-(âˆµtan (90Â° – Î¸) = cot Î¸)

=(cot 55Â°/cot 55Â°) + (cot 78Â°/cot 78Â°) – 1

= 1 + 1 – 1 = 1

Hence,  = 1

### (vi)

Solution:

Given:

-(âˆµ sec (90Â° – Î¸) = cosec Î¸ and sin (90Â° – Î¸) = cos Î¸)

= cosec 20Â°/cosec 20Â° + cos 31Â°/cos 31Â°

= 1 + 1 = 2

Hence,  = 2

### (vii) cosec 31Â° – sec 59Â°

Solution:

Given: cosec 31Â° – sec 59Â°

= cosec 31Â° – sec (90Â°- 31Â°)          -(âˆµ sec (90Â° – Î¸) = cosec Î¸)

= cosec 31Â° – cosec 31Â°

= 0

Hence, cosec 31Â° – sec 59Â° = 0

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

Solution:

Given: (sin 72Â° + cos 18Â°)(sin 72Â° – cos 18Â°)

= (sin 72Â° + cos (90Â° – 72Â°))(sin 72Â° – cos (90Â° – 72Â°))          -(âˆµ sin (90Â° – Î¸) = cos Î¸)

= (sin 72Â° + sin 72Â°)(sin 72Â° – sin 72Â°)

= 0

Hence, (sin 72Â° + cos 18Â°)(sin 72Â° – cos 18Â°) = 0

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

Solution:

Given: sin 35Â°sin 55Â° – cos 35Â°cos 55Â°

= (sin 35Â°sin (90Â°- 35Â°)) – (cos 35Â° cos (90Â° – 35Â°))          -(âˆµ sin (90Â° – Î¸) = cos Î¸ and cos (90Â° – Î¸) = sin Î¸)

= (sin 35Â°cos 35Â°) – (cos 35Â° sin 35Â°)

= 0

Hence, sin 35Â°sin 55Â° – cos 35Â°cos 55Â° = 0

### (x) tan 48Â°tan 23Â°tan 42Â°tan 67Â°

Solution:

Given: tan 48Â°tan 23Â°tan 42Â°tan 67Â°

= tan 48Â° tan 23Â° tan (90Â° – 48Â°) tan (90Â° – 23Â°)           -(âˆµ tan (90Â° – Î¸) = cot Î¸)

= tan 48Â° tan 23Â° cot 48Â° cot 23Â°

= 1

Hence, tan 48Â°tan 23Â°tan 42Â°tan 67Â° = 1

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

Solution:

Given: sec 50Â°sin 40Â° + cos 40Â°cosec 50Â°

= (sec 50Â°sin (90Â° – 50Â°)) + (cos 40Â° cosec(90Â° – 40Â°))          -(âˆµ sin (90Â° – Î¸) = cos Î¸ and cosec (90Â° – Î¸) = sec Î¸)

= (sec 50Â°cos 50Â°) + (cos 40Â° sec 40Â°)

= 1 + 1 = 2

Hence, sec 50Â°sin 40Â° + cos 40Â°cosec 50Â° = 2

### (i) sin 59Â° + cos 56Â°

Solution:

Given: sin 59Â° + cos 56Â°

= sin (90Â° – 31Â°) + cos(90Â° – 34Â°)

= cos 31Â° + sin 34Â°

### (ii) tan 65Â° + cot 49Â°

Solution:

Given: tan 65Â° + cot 49Â°

= tan (90Â° – 25Â°) + cot (90Â° – 31Â°)

= cot 25Â° + tan 31Â°

### (iii) sec 76Â° + cosec 52Â°

Solution:

Given: sec 76Â° + cosec 52Â°

= sec (90Â° – 14Â°) + cosec (90Â° – 38Â°)

= cosec 14Â° + sec 38Â°

### (iv) cos 78Â° + sec 78Â°

Solution:

Given: cos 78Â° + sec 78Â°

= cos (90Â° – 12Â°) + sec (90Â° – 12Â°)

= sin 12Â° + cosec 12Â°

### (v) cosec 54Â° + sin 72Â°

Solution:

Given: cosec 54Â° + sin 72Â°

= cosec (90Â° – 36Â°) + sin (90Â° – 18Â°)

= sec 36Â° + cos 18Â°

### (vi) cot 85Â° + cos 75Â°

Solution:

Given: cot 85Â° + cos 75Â°

= cot (90Â° – 5Â°) + cos (90Â° – 15Â°)

= tan 5Â° + sin 15Â°

### (vii) sin 67Â° + cos 75Â°

Solution:

Given:sin 67Â° + cos 75Â°

= sin (90Â° – 23Â°) + cos (90Â° – 15Â°)

= cos 23Â° + sin 15Â°

### Question 4. Express cos 75Â° + cot 75Â° in terms of angles lying between 0Â° and 30Â°.

Solution:

Given: cos 75Â° + cot 75Â°

= cos (90Â° – 15Â°) + cot (90Â° – 15Â°)

= sin 15Â° + tan 15Â°

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

Solution:

Given: sin 3A = cos(A – 26Â°)

= cos (90Â° – 3A) = cos(A – 26Â°)

Now, 90Â° – 3A = A – 26Â°

= A + 3A = 90Â° + 26Â°

= 4A = 116Â°

= A = 29Â°

Hence, the value of A is 29Â°

### (i) tan (C + A)/2 = cot B/2

Solution:

According to the question

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

So,

A + B + C = 180Â°

C + A = 180Â° – B

Taking LHS

tan (C + A)/2 = tan (180Â° – B)/2

= tan (90Â° – B)/2            -(âˆµ tan (90Â° – Î¸) = cot Î¸)

= cot B/2 = RHS

LHS = RHS

Hence Proved

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

Solution:

According to the question

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

So,

A + B + C = 180Â°

B + C = 180Â° – A

Taking LHS

= sin (B + C)/2 = sin (180Â° – A)/2

= sin (90Â° – A/2)          -(âˆµ sin (90Â° – Î¸) = cos Î¸)

= cos A/2

LHS = RHS

Hence Proved

### (i) tan20Â°tan35Â°tan45Â°tan55Â°tan70Â° = 1

Solution:

We have to prove that tan20Â°tan35Â°tan45Â°tan55Â°tan70Â° = 1

Taking LHS

= tan20Â°tan35Â°tan45Â°tan55Â°tan70Â°

= tan(90Â° âˆ’ 70Â°)tan(90Â° âˆ’ 55Â°)tan45Â°tan55Â°tan70Â°          -(âˆµ tan (90Â° – Î¸) = cot Î¸)

= cot70Â°cot55Â°tan45Â°tan55Â°tan70Â°

= (1/tan70Â°)(1/tan55Â°)tan45Â°tan55Â°tan70Â°                 (âˆµ cot Î¸ = 1/tan Î¸)

= tan45Â°

= 1

LHS = RHS

Hence Proved

### (ii) sin48Â°sec42Â° + cos48Â°cosec42Â° = 2

Solution:

We have to prove that sin48Â°sec42Â° + cos48Â°cosec42Â° = 2

Taking LHS

= sin48Â°sec42Â° + cos48Â°cosec42Â°          (âˆµ sec Î¸ = 1/cos Î¸ and cosec Î¸ = 1/sin Î¸)

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

= sin48Â°/sin48Â° + cos48Â°/cos48Â°

= 1 + 1

= 2

LHS = RHS

Hence Proved

### (iii)  – 2cos70Â° cosec20Â° = 0

Solution:

We have to prove that  – 2cos70Â° cosec20Â° = 0

Taking LHS

= sin 70Â°/cos 20Â° + cosec 20Â°/sec 70Â° – 2cos70Â° cosec20Â°

= sin 70Â°/sin 70Â° + cosec 20Â°/cosec 20Â° – 2cos 70Â°sec 70Â°

= 1 + 1 – 2cos 70Â°/cos 70Â°

= 0

LHS = RHS

Hence Proved

### (iv)  + cos 59Â°cosec31Â° = 2

Solution:

We have to prove that  + cos 59Â°cosec31Â° = 2

Taking LHS

= cos 80Â°/sin 10Â° + cos 59Â°cosec 31Â°

= cos 80Â°/sin (90Â° – 80Â°) + cos 59Â°cosec(90Â°-59Â°)

= cos 80Â°/cos 80Â° + cos 59Â°/cos 59Â°

= 1 + 1 = 2

LHS = RHS

Hence Proved

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