# PROVING TRIGONOMETRIC IDENTITIES WORKSHEET WITH ANSWERS

Proving Trigonometric Identities Worksheet with Answers :

Worksheet given in this section will be much useful for the students who would like to practice solving problems using trigonometric identities.

Before look at the worksheet, if you wish to learn trigonometric identities in detail,

## Proving Trigonometric Identities Worksheet - Questions

Question 1 :

Prove :

(1 - cos2θ) csc2θ  =  1

Question 2 :

Prove :

sec θ √(1 - sin2θ)  =  1

Question 3 :

Prove :

tan θ sin θ + cos θ  =  sec θ

Question 4 :

Prove :

(1 - cos θ)(1 + cos θ)(1 + cot2θ)  =  1

Question 5 :

Prove :

cot θ + tan θ  =  sec θ csc θ

Question 6 :

Prove :

cos θ/(1 - tan θ) + sin θ/(1 - cot θ)  =  sin θ + cos θ

Question 7 :

Prove :

tan4θ + tan2θ  =  sec4θ - sec2θ

Question 8 :

Prove :

√{(sec θ – 1)/(sec θ + 1)}  =  cosec θ - cot θ

Question 9 :

Prove :

(1 - sin A)/(1 + sin A)  =  (sec A - tan A)2

Question 10 :

Prove :

(tan θ + sec θ - 1)/(tan θ - sec θ + 1)  =  (1 + sin θ)/cos θ ## Proving Trigonometric Identities Worksheet - Answers

Question 1 :

Prove :

(1 - cos2θ) csc2θ  =  1

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

A  =  (1 - cos2θ) csc2θ

Because sin2θ + cos2θ  =  1, we have

sin2θ  =  1 - cos2θ

Then,

A  =  sin2θ  csc2θ

A  =  sin2θ  (1/sin2θ)

A  =  sin2θ /sin2θ

A  =  1

A  =  B  (Proved)

Question 2 :

Prove :

sec θ √(1 - sin2θ)  =  1

Let A  =  sec θ √(1 - sin2θ)  and B  =  1.

A  =  sec θ √(1 - sin2θ)

Because sin2θ + cos2θ  =  1, we have

cos2θ  =  1 - sin2θ

Then,

A  =  sec θ √cos2θ

A  =  sec θ  cos θ

A  =  sec θ  (1/sec θ)

A  =  sec θ / sec θ

A  =  1

A  =  B  (Proved)

Question 3 :

Prove :

tan θ sin θ + cos θ  =  sec θ

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

A  =  tan θ sin θ + cos θ

A  =  (sin θ/cos θ)  sin θ + cos θ

A  =  (sin2θ/cos θ) + cos θ

A  =  (sin2θ/cos θ) + (cos2θ/cosθ)

A  =  (sin2θ + cos2θ) / cos θ

A  =  1 / cos θ

A  =  sec θ

A  =  B  (Proved)

Question 4 :

Prove :

(1 - cos θ)(1 + cos θ)(1 + cot2θ)  =  1

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)

Because sin2θ + cos2θ  =  1, we have

sin2θ  =  1 - cos2θ

Then,

A  =  sin2θ  (1 + cot2θ)

A  =  sin2θ  + sin2θ  cot2θ

A  =  sin2θ  + sin2θ  (cos2θ/sin2θ)

A  =  sin2θ + cos2θ

A  =  1

A  =  B  (Proved)

Question 5 :

Prove :

cot θ + tan θ  =  sec θ csc θ

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

A  =  cot θ + tan θ

A  =  (cos θ/sin θ) + (sin θ/cos θ)

A  =  (cos2θ/sin θ cos θ) + (sin2θ/sin θ cos θ)

A  =  (cos2θ + sin2θ) / sin θ cos θ

A  =  1 / sin θ cos θ

A  =  (1/cos θ)  (1/sin θ)

A  =  sec θ csc θ

A  =  B  (Proved)

Question 6 :

Prove :

cos θ/(1 - tan θ) + sin θ/(1 - cot θ)  =  sin θ + cos θ

Let A  =  cos θ/(1 - tan θ) + sin θ/(1 - cot θ)  and

B  =  sin θ + cos θ

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

A  =  cos2θ/(cos θ - sin θ) + sin2θ/(sin θ - cos θ)

A  =  cos2θ/(cos θ - sin θ) - sin2θ/(cos θ - sin θ)

A  =  (cos2θ - sin2θ) / (cos θ - sin θ)

A  =  [(cos θ + sin θ)(cos θ - sin θ)] / (cos θ - sin θ)

A  =  (cos θ + sin θ)

A  =  B  (Proved)

Question 7 :

Prove :

tan4θ + tan2θ  =  sec4θ - sec2θ

Let A  =  tan4θ + tan2θ  and B  =  sec4θ + sec2θ.

A  =  tan4θ + tan2θ

A  =  tan2θ (tan2θ + 1)

We know that,

tan2θ  =  sec2θ - 1

tan2θ + 1  =  sec2θ

Then,

A  =  (sec2θ - 1)(sec2θ)

A  =  sec4θ - sec2θ

A  =  B  (Proved)

Question 8 :

Prove :

√{(sec θ – 1)/(sec θ + 1)}  =  cosec θ - cot θ

Let A  =  √{(sec θ – 1)/(sec θ + 1)} and B  =  cosec θ - cot θ.

A  =  √{(sec θ – 1)/(sec θ + 1)}

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

A  =  √{(sec θ - 1)/ (sec2θ - 1)}

A  =  √{(sec θ - 1)/ tan2θ}

A  =  (sec θ – 1)/tan θ

A  =  (sec θ/tan θ) – (1/tan θ)

A  =  {(1/cos θ)/(sin θ/cos θ)} - cot θ

A  =  {(1/cos θ)  (cos θ/sin θ)} - cot θ

A  =  (1/sin θ) - cot θ

A  =  cosec θ - cot θ

A  =  B  (Proved)

Question 9 :

Prove :

(1 - sin A)/(1 + sin A)  =  (sec A - tan A)2

Let A  =  (1 - sin A)/(1 + sin A) and B  =  (sec A - tan A)2.

A  =  (1 - sin A) / (1 + sin A)

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

A  =  (1 - sin A)/ (1 - sin2A)

A  =  (1 - sin A)/ (cos2A)

A  =  (1 - sin A)/ (cos A)2

A  =  {(1 - sin A) / cos A}2

A  =  {(1/cos A) - (sin A/cos A)}2

A  =  (sec A – tan A)2

A  =  B  (Proved)

Question 10 :

Prove :

(tan θ + sec θ - 1)/(tan θ - sec θ + 1)  =  (1 + sin θ)/cos θ

Let A  =  (tan θ + sec θ - 1)/(tan θ - sec θ + 1)  and

B  =  (1 + sin θ)/cos θ.

A  =  (tan θ + sec θ - 1)/(tan θ - sec θ + 1)

A  =  [(tan θ + sec θ) - (sec2θ - tan2θ)]/(tan θ - sec θ + 1)

A  =  {(tan θ + sec θ) (1 - sec θ + tan θ)}/(tan θ - sec θ + 1)

A  =  {(tan θ + sec θ) (tan θ - sec θ + 1)}/(tan θ - sec θ + 1)

A  =  tan θ + sec θ

A  =  (sin θ/cos θ) + (1/cos θ)

A  =  (sin θ + 1)/cos θ

A  =  (1 + sin θ)/cos θ

A  =  B,   (Proved) After having gone through the stuff given above, we hope that the students would have understood how to solve problems using trigonometric identities.

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