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.

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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

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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

 

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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˚

 

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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˚