PROBLEMS ON TRIGONOMETRIC IDENTITIES WITH SOLUTIONS

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Problem 1 : 

Prove : 

(1 - cos2θ) csc2θ  =  1

Solution :

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)

Problem 2 : 

Prove :

sec θ √(1 - sin2θ)  =  1

Solution :

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)

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  =  (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)

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)

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)

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  =  (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)

Problem 6 : 

Prove :

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

Solution :

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)

Problem 7 : 

Prove :

tan4θ + tan2θ  =  sec4θ - sec2θ

Solution :

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)

Problem 8 : 

Prove :

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

Solution :

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)

Problem 9 : 

Prove :

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

Solution :

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)

Problem 10 : 

Prove :

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

Solution :

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)

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