# TRIGONOMETRIC IDENTITIES WORKSHEET

Prove each of the following.

Problem 1 :

(1 - cos2θ) ⋅ csc2θ = 1

Problem 2 :

Problem 3 :

tanθ ⋅ sinθ + cosθ = secθ

Problem 4 :

(1 - cosθ) ⋅ (1 + cosθ) ⋅ (1 + cot2θ) = 1

Problem 5 :

cotθ + tanθ = secθ ⋅ cscθ

Problem 6 :

Problem 7 :

tan4θ + tan2θ = sec4θ - sec2θ

Problem 8 :

Problem 9 :

Problem 10 :

(1 - cos2θ) ⋅ csc2θ = 1

Let A = (1 - cos2θ)  csc2θ  and B = 1.

Since sin2θ + cos2θ = 1, we have sin2θ = 1 - cos2θ.

Then, we have

A = sin2θ  csc2θ

A = 1

A = B  Proved

Since sin2θ + cos2θ = 1, we have cos2θ = 1 - sin2θ.

A = 1

A = B  Proved

tanθ ⋅ sinθ + cosθ = secθ

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

A = tanθ  sin θ + cos θ

A = secθ

A = B  Proved

(1 - cosθ) ⋅ (1 + cosθ) ⋅ (1 + cot2θ) = 1

Let A = (1 - cosθ)  (1 + cosθ)  (1 + cot2θ) and B = 1.

A = (1 - cosθ)  (1 + cosθ)  (1 + cot2θ)

A = (1 - cos2θ)  (1 + cot2θ)

Since, sin2θ + cos2θ = 1, we have sin2θ = 1 - cos2θ.

A = sin2θ  (1 + cot2θ)

A = sin2θ  + cos2θ

A = 1

A = B  Proved

cotθ + tanθ = secθ ⋅ cscθ

Let A = cotθ + tanθ and B = secθ  cscθ.

A = cotθ + tanθ

Since cos2θ + sin2θ = 1, we have

A = secθ  cscθ

A = B  Proved

A = cosθ + sinθ

A = B  Proved

tan4θ + tan2θ = sec4θ - sec2θ

Let A = tan2θ + tan2θ and B = sec2θ - sec2θ.

A = tan2θ (tan2θ + 1)

A = (sec2θ - 1)(tan2θ + 1)

Since tan2θ = sec2θ – 1, we have

A = (sec2θ - 1) ⋅ sec2θ

Since tan2θ + 1 = sec2θ, we have

A = sec4θ - sec2θ

A = B  Proved

Since sec2θ = 1 + tan2θ, we have sec2θ - 1 = tan2θ

A = cscθ - cotθ

A = B  Proved

A = (secA – tanA)2

A = B  Proved

Since sec2θ - tan2θ = 1, we have

A = B  Proved

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