PROBLEMS ON TRIGONOMETRIC IDENTITIES WITH SOLUTIONS

Problem 1 : 

Prove : 

(1 - cosθ)cscθ = 1

Solution :

Let A = (1 - cosθ)cscθ and B = 1.

A = (1 - cosθ)cscθ

Since sinθ + cosθ = 1, we have  

sinθ = 1 - cosθ

Then, 

A = sinθ  cscθ

A = 1

A = B  (Proved)

Problem 2 : 

Prove :

Solution :



Since sinθ + cosθ = 1, we have  

cosθ = 1 - sinθ

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 sinθ + cosθ = 1, we have  

cosθ = 1 - sinθ

Then, 

A = sin2θ  (1 + cot2θ)

A = sin2θ  + sin2θ  cot2θ

A = sinθ + cosθ

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 :



center>

A = sin θ + cos θ 

A = B  (Proved)

Problem 7 : 

Prove :

tanθ + tanθ = secθ - secθ

Solution :

Let A = tanθ + tanθ and B = secθ + secθ.

A = tanθ + tanθ

A = tanθ (tanθ + 1)

We know that,

tanθ = secθ - 1

tanθ + 1 = secθ 

Then, 

A = (secθ - 1)(secθ)

A = secθ - secθ

A = B  (Proved)

Problem 8 : 

Prove :

Solution :



A = B  (Proved)

Problem 9 : 

Prove :

Solution :



A = (sec θtan θ)2

A = B  (Proved)

Problem 10 : 

Prove :

Solution :



A = B   (Proved)

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