If (cos²θ - 1)(1 + tan²θ) + 2tan²θ = 1, 0ᵒ ≤ θ ≤ 90ᵒ then θ is:

If (cos²θ - 1)(1 + tan²θ) + 2tan²θ = 1, 0ᵒ ≤ θ ≤ 90ᵒ then…

If (cos²θ - 1)(1 + tan²θ) + 2tan²θ = 1, 0^o ≤ θ ≤ 90^o then θ is:

Show answer & explanation

Correct answer: B

Explanation

The correct answer is Option 2 i.e. 45^o.

Understanding/Application

(cos²θ - 1)(1 + tan²θ) + 2tan²θ = 1

(- sin²θ) (sec²θ) + 2tan²θ = 1

(- sin²θ/cos²θ) + 2tan²θ = 1

- tan²θ + 2tan²θ = 1

tan²θ = 1

For 0^o ≤ θ ≤ 90^o

θ = 45^o