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Class 10 RD Sharma Solutions – Chapter 5 Trigonometric Ratios – Exercise 5.3 | Set 1
  • Last Updated : 03 Mar, 2021
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Question 1. Evaluate the following:

(i) sin 20°/cos 70°

Solution:

Given: sin 20°/cos 70°

= sin(90° − 70°)/cos 70°

= cos 70°/cos 70°          -(∵ sin (90° – θ) = cos θ)

= 1



Hence, sin 20°/cos 70° = 1

(ii) cos 19°/sin 71°

Solution:

Given: cos 19°/sin 71°

= cos(90° − 71°)/sin 71°

= sin 71°/sin 71°             -(∵ cos (90° – θ) = sin θ)

= 1

Hence, cos 19°/sin 71° = 1                

(iii) sin 21°/cos 69°

Solution:

Given: sin 21°/cos 69°

= sin(90° − 69°)/cos 69°

= cos 69°/cos 69°            -(∵ sin (90° – θ) = cos θ)

= 1  

Hence, sin 21°/cos 69° = 1                                      

(iv) tan 10°/cot 80°

Solution:

Given: tan 10°/cot 80°

= tan(90° − 80°)/cot 80°

= cot 80°/cot 80°            -(∵ tan (90° – θ) = cot θ)

= 1                                



Hence, tan 10°/cot 80° = 1                      

(v) sec 11°/cosec 79°

Solution:

Given: sec 11°/cosec 79°

= sec(90° − 79°)/cosec 79°

= cosec 79°/cosec 79°          -(∵ sec (90° – θ) = cosec θ)

= 1 

Hence, sec 11°/cosec 79° = 1                

Question 2. Evaluate the following:

 (i) (\frac{sin 49°}{cos 41°})2 + (\frac{cos 41°}{sin 49°})2     

Solution:

Given: (\frac{sin 49°}{cos 41°})2 + (\frac{cos 41°}{sin 49°})

(\frac{sin (90°-41°)}{cos 41°})^2+(\frac{cos 41°}{sin (90°-41°)})^2          -(∵ sin (90° – θ) = cos θ)

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

= 1 + 1 = 2

Hence, (\frac{sin 49°}{cos 41°})2 + (\frac{cos 41°}{sin 49°})2 = 2     

(ii) cos 48° – sin 42°

Solution:

Given: cos 48° – sin 42°

= cos 48° – sin (90°- 48°)           -(∵ sin (90° – θ) = cos θ)

= cos 48° – cos 48°  

= 0

Hence, cos 48° – sin 42° = 0

(iii)  \frac{cot 40°}{tan 50°}-\frac{1}{2}\frac{cos 35°}{sin 55°}      

Solution:

Given:  \frac{cot 40°}{tan 50°}-\frac{1}{2}\frac{cos 35°}{sin 55°}

\frac{cot(90°-50°)}{tan 50°}-\frac{1}{2}(\frac{cos 35°}{sin (90°-35°)})        -(∵ sin (90° – θ) = cos θ and cot (90° – θ) = tan θ)

\frac{tan 50°}{tan 50°}-\frac{1}{2}(\frac{cos 35°}{cos 35°})

= 1 – 1/2 = 1/2

Hence, \frac{cot 40°}{tan 50°}-\frac{1}{2}\frac{cos 35°}{sin 55°} = 1/2

(iv) (\frac{sin 27°}{cos 63°})2– (\frac{cos 63°}{sin 27°})2      

Solution:

Given: (\frac{sin 27°}{cos 63°})2– (\frac{cos 63°}{sin 27°})2  

(\frac{sin (90°-63°)}{cos 63°})^2-(\frac{cos 63°}{sin (90°-63°)})^2             -(∵ sin (90° – θ) = cos θ)

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

= 1 – 1 = 0

Hence, (\frac{sin 27°}{cos 63°})2– (\frac{cos 63°}{sin 27°})2 = 0

(v) \frac{tan 35°}{cot 55°}+\frac{cot 78°}{tan 12°}-1

Solution:

Given: \frac{tan 35°}{cot 55°}+\frac{cot 78°}{tan 12°}-1

\frac{tan (90°-55°)}{cot 55°}+\frac{cot 78°}{tan (90°-78°)}-1          -(∵tan (90° – θ) = cot θ)

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

= 1 + 1 – 1 = 1

Hence, \frac{tan 35°}{cot 55°}+\frac{cot 78°}{tan 12°}-1 = 1

(vi) \frac{sec 70°}{cosec 20°}+\frac{sin 59°}{cos 31°}

Solution:

Given: \frac{sec 70°}{cosec 20°}+\frac{sin 59°}{cos 31°}

\frac{sec(90°−20°)}{cosec 20°}+\frac{sin(90°−31°)}{cos 31°}          -(∵ sec (90° – θ) = cosec θ and sin (90° – θ) = cos θ) 

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

= 1 + 1 = 2

Hence, \frac{sec 70°}{cosec 20°}+\frac{sin 59°}{cos 31°} = 2   

 (vii) cosec 31° – sec 59°

Solution:

Given: cosec 31° – sec 59°

= cosec 31° – sec (90°- 31°)          -(∵ sec (90° – θ) = cosec θ)

= cosec 31° – cosec 31°  

= 0

Hence, cosec 31° – sec 59° = 0

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

Solution:

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

= (sin 72° + cos (90° – 72°))(sin 72° – cos (90° – 72°))          -(∵ sin (90° – θ) = cos θ) 

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

= 0

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

(ix) sin 35°sin 55° – cos 35°cos 55°

Solution:

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

= (sin 35°sin (90°- 35°)) – (cos 35° cos (90° – 35°))          -(∵ sin (90° – θ) = cos θ and cos (90° – θ) = sin θ) 

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

= 0

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

(x) tan 48°tan 23°tan 42°tan 67°

Solution:

Given: tan 48°tan 23°tan 42°tan 67°

= tan 48° tan 23° tan (90° – 48°) tan (90° – 23°)           -(∵ tan (90° – θ) = cot θ)

= tan 48° tan 23° cot 48° cot 23°

= 1

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

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

Solution:

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

= (sec 50°sin (90° – 50°)) + (cos 40° cosec(90° – 40°))          -(∵ sin (90° – θ) = cos θ and cosec (90° – θ) = sec θ)  

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

= 1 + 1 = 2

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

Question 3. Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°

(i) sin 59° + cos 56°

Solution:

Given: sin 59° + cos 56°

= sin (90° – 31°) + cos(90° – 34°)

= cos 31° + sin 34°

(ii) tan 65° + cot 49°

Solution:

Given: tan 65° + cot 49°

= tan (90° – 25°) + cot (90° – 31°)

= cot 25° + tan 31°

(iii) sec 76° + cosec 52°

Solution:

Given: sec 76° + cosec 52°

= sec (90° – 14°) + cosec (90° – 38°)

= cosec 14° + sec 38°

(iv) cos 78° + sec 78°

Solution:

Given: cos 78° + sec 78°

= cos (90° – 12°) + sec (90° – 12°)

= sin 12° + cosec 12°

(v) cosec 54° + sin 72°

Solution:

Given: cosec 54° + sin 72°

= cosec (90° – 36°) + sin (90° – 18°)

= sec 36° + cos 18°

(vi) cot 85° + cos 75°

Solution:

Given: cot 85° + cos 75°

= cot (90° – 5°) + cos (90° – 15°)

= tan 5° + sin 15°

(vii) sin 67° + cos 75° 

Solution:

Given:sin 67° + cos 75° 

= sin (90° – 23°) + cos (90° – 15°)

= cos 23° + sin 15°

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

Solution:

Given: cos 75° + cot 75°

= cos (90° – 15°) + cot (90° – 15°)

= sin 15° + tan 15°

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

Solution:

Given: sin 3A = cos(A – 26°)

= cos (90° – 3A) = cos(A – 26°)

Now, 90° – 3A = A – 26°

= A + 3A = 90° + 26°

= 4A = 116°

= A = 29°

Hence, the value of A is 29°

Question 6. If A, B, C are the interior angles of a triangle ABC, prove that

(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 = 180°

C + A = 180° – B

Taking LHS 

tan (C + A)/2 = tan (180° – B)/2

= tan (90° – B)/2            -(∵ tan (90° – θ) = cot θ)

= 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 = 180°

B + C = 180° – A

Taking LHS  

= sin (B + C)/2 = sin (180° – A)/2

= sin (90° – A/2)          -(∵ sin (90° – θ) = cos θ)

= cos A/2 

LHS = RHS

Hence Proved

Question 7. Prove that:

(i) tan20°tan35°tan45°tan55°tan70° = 1

Solution:

We have to prove that tan20°tan35°tan45°tan55°tan70° = 1

Taking LHS  

= tan20°tan35°tan45°tan55°tan70°

= tan(90° − 70°)tan(90° − 55°)tan45°tan55°tan70°          -(∵ tan (90° – θ) = cot θ)

= cot70°cot55°tan45°tan55°tan70°

= (1/tan70°)(1/tan55°)tan45°tan55°tan70°                 (∵ cot θ = 1/tan θ)

= tan45°

= 1 

LHS = RHS 

Hence Proved

(ii) sin48°sec42° + cos48°cosec42° = 2

Solution:

We have to prove that sin48°sec42° + cos48°cosec42° = 2

Taking LHS

= sin48°sec42° + cos48°cosec42°          (∵ sec θ = 1/cos θ and cosec θ = 1/sin θ)

\frac{sin48°}{cos(90°−48°)} + \frac{cos48°}{sin(90°−48°)}          -(∵ sin (90° – θ) = cos θ and cos (90° – θ) = sin θ)

= sin48°/sin48° + cos48°/cos48°

= 1 + 1

= 2

LHS = RHS

Hence Proved

(iii) \frac{sin 70°}{cos 20°}+\frac{cosec 20°}{sec 70°} – 2cos70° cosec20° = 0 

Solution:

We have to prove that \frac{sin 70°}{cos 20°}+\frac{cosec 20°}{sec 70°} – 2cos70° cosec20° = 0 

Taking LHS

= sin 70°/cos 20° + cosec 20°/sec 70° – 2cos70° cosec20°

\frac{sin 70°}{cos (90°-70°)} + \frac{cosec 20°}{sec(90°-20°)} - 2cos 70°cosec (90°-70°)

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

= 1 + 1 – 2cos 70°/cos 70°

= 0 

LHS = RHS 

Hence Proved

(iv) \frac{cos 80°}{sin 10°} + cos 59°cosec31° = 2

Solution:

We have to prove that \frac{cos 80°}{sin 10°} + cos 59°cosec31° = 2

Taking LHS 

= cos 80°/sin 10° + cos 59°cosec 31°

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

= cos 80°/cos 80° + cos 59°/cos 59°

= 1 + 1 = 2 

LHS = RHS 

Hence Proved

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