Previous Year Questions

Let us factor the terms in the numerator and denominator:

(sin θ - 2 sin³θ ) = sin θ (1- 2 sin² θ)

(2 cos³ θ -cos θ)=cos θ(2cos² θ -1)

Now;

[(sin θ (1-2 sin² θ))/(cos θ (2 ccs² θ -1))]³x(1/tan θ) -sec² θ

Now, use the identity tan θ = sin θ / cos θ, and simplify:

(sin³ θ (1-2 sin² θ)³)/(cos³ θ (2cos² θ -1)³) X (cos θ /sin θ)-sec²θ

After canceling sin θ and cos θ terms:

(sin²θ (1 -2 sin² θ)³)/(cos² θ (2 cos² θ -1)³) -sec² θ

We know that 2 cos² θ - 1 = cos 2 θ, so the expression becomes;

(sin² θ (1 -2 sin² θ)³)/(cos² θ (cos 2 θ)³) - sec² θ

Using the identity cos 2 θ = 1-2 sin² θ:

(sin² θ (cos 2 θ)³)/ (cos² θ (cos 2 θ)³)- sec² θ

Cancel out the term (cos 2 θ)³

Now the expression become:

(sin2 θ / cos² θ)-sec² θ

tan² θ - sec² θ

using identity: 1 + tan² θ = sec² θ

tan² θ -sec² θ=-1

option (d) is the correct answer.