# TRIGONOMETRIC IDENTITIES WORKSHEET

Trigonometric Identities Worksheet :

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 would like to know the different trigonometric identities,

## Trigonometric Identities Worksheet - Problems

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 :

Tanθ + Tan²θ  =  Secθ - 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 ## Trigonometric Identities Worksheet - Solutions

Problem 1 :

Prove :

(1 - Cos²θ)  Csc²θ  =  1

Solution :

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

Problem 2 :

Prove :

Secθ  √(1 - Sin²θ)  =  1

Solution :

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

Problem 3 :

Prove :

Tanθ  Sinθ + Cosθ  =  Secθ

Solution :

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

A  =  tanθ  sin θ + cos θ

A  =  (sin θ / cosθ)  sinθ + cosθ

A  =  (sin²θ / cosθ) + cosθ

A  =  (sin²θ + cos²θ) / cosθ

A  =  1 / cosθ

A  =  secθ

A  =  B  Proved

Problem 4 :

Prove :

(1 - Cosθ)  (1 + Cosθ)  (1 + Cot²θ)  =  1

Solution :

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  + sin² θ  cot²θ

A  =  sin²θ  + sin²θ  (cos²θ / sin²θ)

A  =  sin²θ  + cos²θ

A  =  1

A  =  B  Proved

Problem 5 :

Prove :

Cotθ + Tanθ  =  Secθ  Cscθ

Solution :

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

Problem 6 :

Prove :

Cosθ / (1 - Tanθ) + Sinθ / (1 - Cotθ)  =  Sinθ + Cosθ

Solution :

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

B  =  sinθ + cosθ

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

A  =  cos θ/{(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

Problem 7 :

Prove :

Tan⁴θ + Tan²θ  =  Sec⁴θ - Sec²θ

Solution :

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

A  =  tan²θ (tan²θ + 1)

A  =  (sec²θ - 1)(tan²θ + 1)

[Bcause, tan²θ  =  sec²θ – 1]

A  =  (sec²θ - 1) ⋅ sec²θ

[Because, tan²θ + 1  =  sec²θ]

A  =  sec⁴θ - sec²θ

A  =  B  Proved

Problem 8 :

Prove :

√{(Secθ – 1) / (Secθ + 1)}  =  Cscθ - Cotθ

Solution :

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 sec² θ - 1 = tan² θ]

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

Problem 9 :

Prove :

(1 - SinA) / (1 + SinA)  =  (SecA - TanA)²

Solution :

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

Problem 10 :

Prove :

(Tanθ + Secθ - 1) / (Tanθ - Secθ + 1)  =  (1 + Sinθ) / Cosθ

Solution :

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

B  =  (1 + sinθ) / cosθ

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

A  =  [(tanθ + secθ) - (sec
²θ - 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 After having gone through the stuff given above, we hope that the students would have understood how to solve proving problems using trigonometric identities.

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