Proving trigonometric identities worksheet with answers pdf

Problem 1 : 

Prove :

 (1 - cos2θ) ⋅ csc2θ = 1

Problem 2 :

Prove :

secθ ⋅ √(1 - sin2θ) = 1

Problem 3 :

Prove :

tanθ ⋅ sinθ + cosθ = secθ

Problem 4 :

Prove :

 (1 - cosθ) ⋅ (1 + cosθ) ⋅ (1 + cot2θ) = 1

Problem 5 : 

Prove :

cotθ + tanθ = secθ ⋅ cscθ

Problem 6 : 

Prove :

cosθ/(1 - tanθ) + sinθ/(1 - cotθ) = sinθ + cosθ 

Problem 7 :

Prove :

tan4θ + tan2θ = sec4θ - sec2θ 

Problem 8 : 

Prove :

√{(secθ – 1)/(secθ + 1)} = cscθ - cotθ 

Problem 9 :

Prove :

(1 - sinA)/(1 + sinA) = (secA - tanA)2

Problem 10 :

Prove :

(tanθ + secθ - 1)/(tanθ - secθ + 1)  =  (1 + sinθ)/cosθ

Answers

1. Answer :

 (1 - cos2θ) ⋅ csc2θ = 1

Let A = (1 - cos2θ) ⋅ csc2θ  and B = 1.

Because,  sin2θ + cos2θ = 1, we have  sin2θ = 1 - cos2θ

Then, we have

A = sin2θ ⋅ csc2θ

We know that csc2θ = 1/sin2θ

A = sin2θ ⋅ 1/sin2θ

A = 1

A = B  Proved

2. Answer :

secθ ⋅ √(1 - sin2θ) = 1

Let A = secθ ⋅ √(1 - sin2θ) and B = 1.

Because sin2θ + cos2θ = 1, we have cos2θ = 1 - sin2θ.

A = secθ ⋅ √cos2θ

A = secθ ⋅ cosθ

A = secθ ⋅ 1/secθ

A = 1

A = B  Proved

3. Answer :

tanθ ⋅ sinθ + cosθ = secθ

Let A = tanθ ⋅ sin θ + cos θ  and B = sec θ.

A = tanθ ⋅ sin θ + cos θ

A = (sinθ/cosθ) ⋅ sinθ + cosθ

A = (sin2θ/cosθ) + cosθ

A = (sin2θ + cos2θ)/cosθ

A = 1/cosθ

A = secθ

A = B  Proved

4. Answer :

 (1 - cosθ) ⋅ (1 + cosθ) ⋅ (1 + cot2θ) = 1

Let A = (1 - cosθ) ⋅ (1 + cosθ) ⋅ (1 + cot2θ) and B = 1.

A = (1 - cosθ) ⋅ (1 + cosθ) ⋅ (1 + cot2θ)

A = (1 - cos2θ) ⋅ (1 + cot2θ)

Because, sin2θ + cos2θ = 1, we have sin2θ = 1 - cos2θ.

A = sin2θ ⋅ (1 + cot2θ)

A = sin2θ  + sin2θ ⋅ cos2θ

A = sin2θ + sin2θ ⋅ (cos2θ/sin2θ)

A = sin2θ  + cos2θ

A = 1

A = B  Proved

5. Answer :

cotθ + tanθ = secθ ⋅ cscθ

Let A = cotθ + tanθ and B = secθ ⋅ cscθ.

A = cotθ + tanθ

A = cosθ/sinθ + sinθ/cosθ

A = (cos2θ + sin2θ)/(sinθ ⋅ cosθ)

(Because, cos2θ + sin2θ = 1)

A = 1/(sinθ ⋅ cosθ)

A = (1/cos θ) ⋅ (1/sin θ)

A = secθ ⋅ cscθ

A = B  Proved

6. Answer :

cosθ/(1 - tanθ) + sinθ/(1 - cotθ) = sinθ + cosθ 

Let A = cosθ/(1 - tanθ) + sinθ/(1 - cotθ)  and

B = sinθ + cosθ.

A = cosθ/{1 - (sinθ/cos θ)} + sinθ/{1 - (cosθ/sinθ)}

A = cosθ/{(cosθ - sinθ)/cosθ} + sinθ/{(sinθ - cosθ/sinθ)} 

A = cos2θ/(cosθ - sinθ) + sin2θ/(sinθ - cosθ) 

A = cos2θ/(cosθ - sinθ) - sin2θ/(cosθ - sinθ) 

A = (cos2θ - sin2θ)/(cosθ - sinθ) 

A = [(cosθ + sinθ)(cosθ - sinθ)]/(cosθ - sinθ) 

A = cosθ + sinθ 

A = B  Proved

7. Answer :

tan4θ + tan2θ = sec4θ - sec2θ

Let A = tan2θ + tan2θ  and B = sec2θ - sec2θ. 

A = tan2θ (tan2θ + 1) 

A = (sec2θ - 1)(tan2θ + 1)

[Because tan2θ = sec2θ – 1] 

A = (sec2θ - 1) ⋅ sec2θ  

[Because, tan2θ + 1 = sec2θ] 

A = sec4θ - sec2θ

A = B  Proved

8. Answer :

√{(secθ – 1)/(secθ + 1)} = cscθ - cotθ

Let A = √{(secθ – 1)/(secθ + 1)} and B = cosecθ - cotθ.

A = √{(secθ – 1)/(secθ + 1)} 

A = √[{(secθ - 1) (secθ - 1)}/{(secθ + 1) (secθ - 1)}]

[Multiplying numerator and denominator by (secθ - l) inside radical sign]

A = √{(secθ - 1)2/(sec2θ - 1)} 

A = √{(secθ -1)2/tan2θ}

[Because, sec2 θ = 1 + tan2 θ ⇒ sec2θ - 1 = tan2θ] 

A = (secθ – 1)/tanθ

A = (secθ/tanθ) – (1/tanθ) 

A = {(1/cosθ)/(sinθ/cos θ)} - cotθ 

A = {(1/cosθ) ⋅ (cosθ/sinθ)} - cotθ

A = (1/sinθ) - cotθ

A = cosecθ - cotθ

A = B  Proved

9. Answer :

(1 - sinA)/(1 + sinA) = (secA - tanA)2

Let A = (1 - sinA)/(1 + sinA) and B = (secA - tanA)2.

A = (1 - sinA)/(1 + sinA)

A = (1 - sinA)2/(1 - sinA)(1 + sinA)

[Multiply both numerator and denominator by (1 - sin A) 

A = (1 - sinA)2/(1 - sin2A) 

A = (1 - sinA)2/(cos2A)

[Because, sin2θ + cos2θ  =  1  ⇒  cos2θ  =  1 - sin2θ] 

A = {(1 - sinA)/cosA}2

A = (1/cosA - sinA/cosA)2

A = (secA – tanA)2

A = B  Proved

10. Answer :

(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/cosθ

Let A = (tanθ + secθ - 1)/(tanθ - secθ + 1)  and

B = (1 + sinθ)/cosθ.

A = (tanθ + secθ - 1)/(tanθ - secθ + 1)

A = [(tanθ + secθ) - (sec2θ - tan2θ)]/(tanθ - secθ + 1)

[Because sec2θ - tan2θ = 1]

A = {(tanθ + secθ) - (secθ + tanθ)(secθ - tanθ)}/(tanθ - secθ + 1)

A = {(tanθ + secθ) (1 - secθ + tanθ)}/(tanθ - secθ + 1) 

A = {(tanθ + secθ) (tanθ - secθ + 1)}/(tanθ - secθ + 1) 

A = tanθ + secθ

A = (sinθ/cosθ) + (1/cosθ) 

A = (sinθ + 1)/cosθ

A = (1 + sinθ)/cosθ 

A = B  Proved

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