PROBLEMS ON TRIGONOMETRIC IDENTITIES WITH SOLUTIONS

About "Problems on trigonometric identities with solutions"

Problems on trigonometric identities with solutions are much useful to the kids who would like to practice problems on trigonometric identities.

Before we look at the problems on trigonometric identities, let us have a look on some important trigonometric identities.

Some important trigonometric Identities

Let us see the reciprocal trigonometric identities

sinθ = 1 / cscθ

cscθ = 1 / sinθ

cosθ = 1 / secθ

secθ = 1 / cos θ

tanθ = 1 / cot θ

cotθ = 1 / tan θ

sin²θ + cos²θ = 1

sin²θ = 1 - cos²θ

cos²θ = 1 - sin²θ

sec²θ - tan²θ = 1

sec²θ = 1 + tan²θ

tan²θ = sec²θ - 1

csc²θ - cot²θ = 1

csc²θ = 1 + cot²θ

cot²θ = csc²θ - 1

Problems on trigonometric identities with solutions

Problem 1 :

Prove : (1 - cos² θ) csc² θ = 1

Solution :

Let A = (1 - cos² θ) csc² θ and

B = 1

A = (1 - cos² θ) csc² θ

Since sin² θ + cos² θ = 1, we have sin² θ = 1 - cos² θ

A = sin² θ x csc² θ

We know that csc² θ = 1/sin² θ

A = sin² θ x 1/sin² θ

A = 1

A = B Proved

Let us look at the next problem on "Problems on trigonometric identities with solutions"

Problem 2 :

Prove : sec θ √(1 - sin ² θ) = 1

Solution :

Let A = sec θ √(1 - sin ² θ) and

B = 1

A = sec θ √(1 - sin ² θ)

Since sin² θ + cos² θ = 1, we have cos² θ = 1 - sin² θ

A = sec θ √cos ² θ

A = sec θ x cos θ

A = sec θ x 1/sec θ

A = 1

A = B Proved

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Problem 3 :

Prove : tan θsin θ + cos θ = sec θ

Solution :

Let A = tan θsin θ + cos θ and

B = sec θ

A = tan θsin θ + cos θ

A = (sin θ/cos θ)sin θ + cos θ

A = (sin² θ/cos θ) + cos θ

A = (sin² θ + cos² θ) / cos θ

A = 1 / cos θ

A = sec θ

A = B Proved

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Problem 4 :

Prove : (1 - cos θ)(1 + cos θ)(1 + cot² θ) = 1

Solution :

Let A = (1 - cos θ)(1 + cos θ)(1 + cot² θ) = 1

B = 1

A = (1 - cos θ)(1 + cos θ)(1 + cot² θ)

A = (1 - cos²)(1 + cot² θ)

Since sin² θ + cos² θ = 1, we have sin² θ = 1 - cos² θ

A = sin² θ x (1 + cot² θ)

A = sin² θ + sin² θ x cot² θ

A = sin² θ + sin² θ x (cos² θ/sin² θ)

A = sin² θ + cos² θ

A = 1

A = B Proved

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Problem 5 :

Prove : cot θ + tan θ = sec θcsc 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/cosθ)

A = (cos² θ + sin² θ) / sin θcosθ

(Because, cos² θ + sin² θ = 1 )

A = 1 / sin θcos θ

A = (1/cos θ) x (1/sin θ)

A = sec θcscθ

A = B Proved

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

A = cos² θ/(cos θ - sin θ) + sin² θ/(sin θ - cos θ)

A = cos² θ/(cos θ - sin θ) - sin² θ/(cos θ - sin θ)

A = (cos² θ - sin² θ)/(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

Let us look at the next problem on "Problems on trigonometric identities with solutions"

Problem 7 :

Prove : tan⁴ θ + tan² θ = sec⁴ θ - sec² θ

Solution :

Let A = tan⁴ θ + tan² θ and

B = sec⁴ θ - sec² θ

A = tan² θ (tan² θ + 1)

A = (sec² θ - 1) (tan² θ + 1) [since, tan² θ = sec² θ – 1]

A = (sec² θ - 1) sec² θ [since, tan² θ + 1 = sec² θ]

A = sec⁴ θ - sec² θ

A = B Proved

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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)}]

[multiplying numerator and denominator by (sec θ - l) under radical sign]

A = √{(sec θ - 1)²/(sec² θ - 1)}

A = √{(sec θ -1)²/tan² θ}

[since, sec² θ = 1 + tan² θ ⇒ sec² θ - 1 = tan² θ]

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

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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)

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

A = (1 - sin A)²/(1 - sin² A)

A = (1 - sin A)²/(cos² A)

[Since sin² θ + cos² θ = 1 ⇒ cos² θ = 1 - sin² θ]

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

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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) - (sec² θ - tan² θ)]/(tan θ - sec θ + 1)

[Since, sec² θ - tan² θ = 1]

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