The value of $\frac{\tan^2 \theta - \sin^2 \theta}{2 + \tan^2 \theta + \cot^2 \theta}$ is:

$\sin^6 \theta$

$\frac{\tan^2 \theta - \sin^2 \theta}{2 + \tan^2 \theta + \cot^2 \theta}$

Put θ = 45º

= \(\frac{tan²45º - sin²45º}{ 2 + tan²45º + cot²45º}\)

= \(\frac{1- 1/2}{ 2 + 1 + 1}\)

= \(\frac{1}{8}\)

Now, by putting θ = 45º in all options one by one . And option 3 satisfied the condition.

sin6θ

= sin645º

= ( \(\frac{1}{√2}\))⁶

= \(\frac{1}{8}\)

Ans :- sin6θ