Trigonometry Exercise 13.4 – Class 10

Trigonometry Exercise 13.4 – Solutions:

  1. Evaluate:
  2. tan65˚/cot25˚
  3. sin18˚/cos72˚

iii. cos48˚- sin42˚

  1. cosec31˚ – sec59˚
  2. cot34˚ – tan56˚
  3. sin36˚/cos54˚sin54˚/cos36˚

vii. sec70˚ sin20˚ – cos70˚cosec20˚

viii. cos213˚ – sin277˚

 

  1. Prove that:
  2. sin35˚ sin55˚ – cos35˚cos55˚ = 0
  3. tan10˚tan15˚tan75˚tan80˚ = 1

iii. cos38˚cos52˚ – sin38˚sin52˚ = 0

 

III. If sin5θ = cos4θ, where 4θ and 5θ are acute angles, find the value of θ

 

  1. If sec4A = cosec(A – 20˚), where 4A is an acute angle, find the value of A.

Trigonometry Exercise 13.4 – Questions:

I. Evaluate:

  1. tan65˚/cot25˚

Solution:

We know,

cot25˚ = tan(90˚ – 25˚) = tan65˚

Therefore,  tan65˚/cot25˚ = tan65˚/tan65˚ =  1

 

 

ii. sin18˚/cos72˚

Solution:

We know,

cos72˚ = sin(90˚ – 72˚) = sin18˚

Therefore,  sin18˚/cos72˚ = sin18˚/sin18˚ =  1

 

iii. cos48˚- sin42˚

Solution:

We know,

cos48˚ = sin(90˚ – 48˚) = sin48˚

Therefore,  cos48˚- sin42˚= cos48˚- cos48˚ =  0

 

iv.  cosec31˚ – sec59˚

Solution:

We know,

cosec31˚ = sec(90˚ – 31˚) = sec59˚

Therefore,  cosec31˚ – sec59˚= cosec31˚ – cosec31˚=  0

 

v. cot34˚ – tan56˚

Solution:

We know,

cot34˚ = tan(90˚ – 34˚) = tan56˚

Therefore,  cot34˚ – tan56˚= cot34˚ – cot34˚=  0

 

vi. sin36˚/cos54˚sin54˚/cos36˚

Solution:

We know,

cos54˚ = sin(90˚ – 54˚) = sin36˚

cos36˚ = sin(90˚ – 36˚) = sin54˚

Therefore,  sin36˚/cos54˚sin54˚/cos36˚ = sin36˚/sin36˚sin54˚/sin54˚=  1 – 1  = 0

 

vii. sec70˚ sin20˚ – cos70˚cosec20˚

Solution:

We know,

sec70˚ = cosec(90˚ – 70˚) = cosec20˚

sec20˚ = cos(90˚ – 20˚) = cos70˚

Therefore,  sec70˚ sin20˚ – cos70˚cosec20˚ = cosec20˚ cos70˚ – cos70˚cosec20˚=  0

 

viii. cos213˚ – sin277˚

Solution:

We know, cos213˚ – sin277˚

cos213˚ = 1 – sin213˚

sin213˚ = cos2(90˚ – 13˚) = sin2(77˚)

Therefore,  cos213˚ – sin277˚ = 1 – sin277˚- sin277˚=  1


II. Prove that:

  1. sin35˚ sin55˚ – cos35˚cos55˚ = 0

Solution:

sin35˚ sin55˚ – cos35˚cos55˚ = 0

cos35˚ = sin(90˚ – 35˚) = sin55˚

cos55˚ = sin(90˚ –  55˚) =  sin35˚

Therefore, sin35˚ sin55˚ – cos35˚cos55˚

= sin35˚ sin55˚ – sin35˚ sin55˚

= 0

 

ii. tan10˚tan15˚tan75˚tan80˚ = 1

Solution:

sin10˚/cos10 . sin15˚/cos15. sin75˚/cos75. sin80˚/cos80 = 1

cos 10˚ = sin(90˚ – 10˚) = sin 80˚

cos 15˚ = sin(90˚ – 15˚) = sin75˚

cos75˚ = sin(90˚ – 75˚)  = sin15˚

cos80˚ = sin(90˚ – 10˚) = sin 10˚

⇒  sin10˚/sin80˚ . sin15˚/sin75˚. sin75˚/sin15˚. sin80˚/sin10˚ = 1

⇒ tan10˚tan15˚tan75˚tan80˚ = 1

 

iii. cos38˚cos52˚ – sin38˚sin52˚ = 0

Solution:

cos38˚cos52˚ – sin38˚sin52˚ = 0

we know,

cos38 = sin(90 – 38) = sin 52

cos52 = sin(90 – 52) = sin38

⇒ cos38˚cos52˚ – sin38˚sin52˚ = sin52˚sin38˚ – sin38˚sin52˚ = 0

 


III. If sin5θ = cos4θ, where 4θ and 5θ are acute angles, find the value of θ

Solution:

Given, sin5θ = cos4θ

We know, cos4θ = sin(90˚ – 4θ)

Since, sin(90˚ – 4θ) = sin5θ

⇒ 90˚ – 4θ = 5θ

⇒ 90˚ = 9θ

⇒ θ = 90˚/9 = 10˚

 


IV. If sec4A = cosec(A – 20˚), where 4A is an acute angle, find the value of A.

Solution:

Given  sec4A = cosec(A – 20˚)

sec4A = cosec(90˚ – 4A)

Since cosec(90˚ – 4A) = cosec(A – 20˚)

90˚ – 4A = A – 20˚

90˚ + 20˚ = A + 4A

110˚ = 5A

A = 110˚/5 = 22˚