Question

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

Answer 1

L.H.S

$\Rightarrow \frac{{\tan }^2\theta }{{(\sec \theta -1)}^2}$

$\Rightarrow \frac{({\sec }^2\theta -1)}{{(\sec \theta -1)}^2}[\because {\sec }^2\theta =1+{\tan }^2\theta ]$

$\Rightarrow \frac{(\sec \theta -1)(\sec \theta +1)}{{(\sec \theta -1)}^2}$

$\Rightarrow \frac{(\sec \theta +1)}{(\sec \theta -1)}$

$\Rightarrow \frac{(\frac{1}{\cos \theta }+1)}{(\frac{1}{\cos \theta }-1)}$

$\Rightarrow \frac{(1+\cos \theta )}{(1-\cos \theta )}$

R.H.S

Hence, proved.