Question

tan square theta -sin square theta=tan square theta sin square theta

Answer 1

Here i have used x instead of theta

To prove : tan²x - sin²x=tan²xsin²x
Proof:
LHS= tan²x - sinx
=sin²x/cos²x - sin²x
Taking L.C.M we get
LHS=(Sin²x-sin²x. cos²x) /cos²x
= sin²x (1 - cos²x) /cos²x
=tan²x. sin²x
(1-cos²x=sin²x)

Answer 2

Hi...☺

Here is your answer...✌

TO PROVE :
$\tan {}^{2} \theta - \sin {}^{2} \theta = \tan {}^{2} \theta \sin {}^{2} \theta$

PROOF :

LHS

$= \tan {}^{2} \theta - \sin {}^{2} \theta \\ \\ = \frac{\sin {}^{2} \theta}{\cos {}^{2} \theta} - \sin {}^{2} \theta\\ \\ = \sin {}^{2} \theta( \frac{1}{ \cos {}^{2} \theta } - 1) \\ \\ = \sin {}^{2} \theta(\sec {}^{2} \theta - 1) \\ \\ = \sin {}^{2} \theta\tan{}^{2} \theta$

= RHS [Proved]