# TRIGONOMETRIC IDENTITIES WORKSHEET

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 :

tan4θ + tan2θ  =  sec4θ - 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 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  + sin2θ  cos2θ

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  =  co
s2θ/(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  =  tan2θ + tan2θ  and B  =  sec2θ - 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 sec2θ - 1 = tan2θ]

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 Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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