(3 cos³ θ - 2 cos θ)/(sin θ - 3 sin³ θ) is equal to:

$\frac{3 \cos^3 \theta - 2 \cos \theta}{\sin \theta - 3 \sin^3 \theta}$ is equal to:

Show answer & explanation

Correct answer: C

Explanation

The correct answer is option 3 i.e. cot θ.

⇒ $\frac{3 \cos^3 \theta - 2 \cos \theta}{\sin \theta - 3 \sin^3 \theta}$

⇒ [cos θ(3cos²θ - 2)]/[sin θ(1 - 3sin²θ)]

We know,

cos²θ = 1 - sin²θ

⇒ [cos θ(3(1 - sin²θ) - 2)]/[sin θ(1 - 3sin²θ)]

⇒ [cos θ(1 - 3sin²θ)]/[sin θ(1 - 3sin²θ)]

⇒ cos θ/sin θ = cot θ

SSC CHSL 2024 PYQ — Question 75