**Trigonometry Exercise 13.4 – Solutions:**

**Evaluate:**^{tan65˚}/_{cot25˚}^{sin18˚}/_{cos72˚}

**iii. cos48˚- sin42˚**

**cosec31˚ – sec59˚****cot34˚ – tan56˚**^{sin36˚}/_{cos54˚}–^{sin54˚}/_{cos36˚}

**vii. sec70˚ sin20˚ – cos70˚cosec20˚**

**viii. cos ^{2}13˚ – sin^{2}77˚**

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

**iii. cos38˚cos52˚ – sin38˚sin52˚ = 0**

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**III. If sin5**θ** = cos4**θ**, where 4**θ** and 5**θ** are acute angles, find the value of θ**

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

**Trigonometry Exercise 13.4 – Questions:**

**I. Evaluate:**

^{tan65˚}/_{cot25˚}

Solution:

We know,

cot25˚ = tan(90˚ – 25˚) = tan65˚

Therefore, ^{tan65˚}/_{cot25˚} = ^{tan65˚}/_{tan65˚} = 1

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^{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. cos ^{2}13˚ – sin^{2}77˚**

Solution:

We know, cos^{2}13˚ – sin^{2}77˚

cos^{2}13˚ = 1 – sin^{2}13˚

sin^{2}13˚ = cos^{2}(90˚ – 13˚) = sin^{2}(77˚)

Therefore, cos^{2}13˚ – sin^{2}77˚ = 1 – sin^{2}77˚- sin^{2}77˚= 1

**II. Prove that:**

**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˚