The value of $\frac{\tan^2 \theta - \sin^2 \theta}{2 + \tan^2 \theta + \cot^2 \theta}$ is:
Answer & explanation
Correct answer: option 3
$\frac{\tan^2 \theta - \sin^2 \theta}{2 + \tan^2 \theta + \cot^2 \theta}$
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