If θ ≠ 45°, ( (sinθ - 2sin³θ)/(2cos³θ - cos³θ) )² + 1 is equal to:

If θ ≠ 45°, ( (sinθ - 2sin³θ)/(2cos³θ - cos³θ) )² + 1 is…

If θ ≠ 45°, $\left( \frac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos^3\theta} \right)^2$ + 1 is equal to:

Show answer & explanation

Correct answer: D

Explanation

The correct answer is option 4 i.e. sec²θ

(1 – 2sin²θ) = (2cos²θ – 1) = cos2θ

And 1 + tan²θ = sec²θ

Now, [(sinθ - 2sin³θ)/(2cos³θ - cosθ)]²+ 1

⇒ (sinθ/cosθ)² [(1 - 2sin²θ/(2cos²θ -1)]²+ 1

⇒ (tan²θ)(cos2θ/cos2θ)² + 1

⇒ tan²θ × 1 + 1

⇒ 1 + tan²θ

⇒ sec²θ