Question

please solve it...please please,........​

Answer 1

Given :-

$\dfrac{- tan^2 \theta }{sec^2 \theta }+ sin^2 \theta$

To prove :-

0

Solution:-

As we know that :-

$tan\theta = \dfrac{sin\theta }{cos\theta }$

And $Sec\theta = \dfrac{1}{cos\theta }$

Now change $tan\theta$ and $sec\theta$

→$\dfrac{\dfrac{-sin^2\theta }{\cos^2\theta}}{\dfrac{1}{cos^2\theta}}+ sin^2 \theta$

→$-sin^2 \theta + sin^2 \theta$

→$= 0$

hence, proved .....