PROBLEMS ON TRIGONOMETRIC IDENTITIES WITH SOLUTIONS

Problem 1 : 

Prove : 

(1 - cos² θ)csc² θ = 1

Solution :

Let A = (1 - cos² θ)csc² θ and B = 1.

A = (1 - cos² θ)csc² θ

Since sin² θ + cos² θ = 1, we have  

sin² θ = 1 - cos² θ

Then, 

A = sin² θ ⋅ csc² θ

A = 1

A = B  (Proved)

Problem 2 : 

Prove :

Solution :

Since sin² θ + cos² θ = 1, we have  

cos² θ = 1 - sin² θ

Then, 

A = 1

A = B  (Proved)

Problem 3 : 

Prove :

tan θ sin θ + cos θ = sec θ

Solution :

Let A = tan θ sin θ + cos θ  and B = sec θ.

A = tan θ sin θ + cos θ

A = sec θ

A = B  (Proved)

Problem 4 : 

Prove :

(1 - cos θ)(1 + cos θ)(1 + cot²θ) = 1

Solution :

Let A = (1 - cos θ)(1 + cos θ)(1 + cot²θ) = 1 and B = 1.

A = (1 - cos θ)(1 + cos θ)(1 + cot²θ)

A = (1 - cos²θ)(1 + cot²θ)

Since sin² θ + cos² θ = 1, we have  

cos² θ = 1 - sin² θ

Then, 

A = sin²θ ⋅ (1 + cot²θ)

A = sin²θ  + sin²θ ⋅ cot²θ

A = sin² θ + cos² θ

A = 1

A = B  (Proved)

Problem 5 : 

Prove :

cot θ + tan θ = sec θ csc θ

Solution :

Let A = cot θ + tan θ and B = sec θ csc θ.

A = cot θ + tan θ

A = csc θ sec θ

A = sec θ csc θ

A = B  (Proved)

Problem 6 : 

Prove :

Solution :

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A = sin θ + cos θ 

A = B  (Proved)

Problem 7 : 

Prove :

tan⁴ θ + tan² θ = sec⁴ θ - sec² θ

Solution :

Let A = tan⁴ θ + tan² θ and B = sec⁴ θ + sec² θ.

A = tan⁴ θ + tan² θ

A = tan² θ (tan² θ + 1)

We know that,

tan² θ = sec² θ - 1

tan² θ + 1 = sec² θ 

Then, 

A = (sec² θ - 1)(sec² θ)

A = sec⁴ θ - sec² θ

A = B  (Proved)

Problem 8 : 

Prove :

Solution :

A = B  (Proved)

Problem 9 : 

Prove :

Solution :

A = (sec θ – tan θ)²

A = B  (Proved)

Problem 10 : 

Prove :

Solution :

A = B   (Proved)

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