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)² / (sec²θ - 1)}

A  =  √{(sec θ - 1)² / 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)² / (1 - sin A) (1 + sin A)

A  =  (1 - sin A)² / (1 - sin²A) 

A  =  (1 - sin A)² / (cos²A)

A  =  (1 - sin A)² / (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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