Question

If $\sec \theta +\tan \theta =x$, show that $\frac{x^2-1}{x^2+1}=\sin \theta$

Answer 1

Given:- $x=\sec \theta +\tan \theta$

To prove:- $\frac{x^2-1}{x^2+1}=\sin \theta$

Proof:-

$x=\sec \theta +\tan \theta$

$\Rightarrow x=\frac{1+\sin \theta }{\cos \theta }$

Squaring both sides, we get

$\Rightarrow x^2=\frac{{(1+\sin \theta )}^2}{{\cos }^2\theta }$

$\Rightarrow x^2=\frac{{(1+\sin \theta )}^2}{1-{\sin }^2\theta }$

$\Rightarrow x^2=\frac{{(1+\sin \theta )}^2}{(1+\sin \theta )(1-\sin \theta )}$

$\Rightarrow x^2=\frac{(1+\sin \theta )}{1-\sin \theta }$

Therefore,

$\frac{x^2-1}{x^2+1}$

$=\frac{\frac{(1+\sin \theta )}{1-\sin \theta }-1}{\frac{(1+\sin \theta )}{1-\sin \theta }+1}$

$=\frac{1+\sin \theta -1+\sin \theta }{1+\sin \theta +1-\sin \theta }$

$=\frac{2\sin \theta }{2}=\sin \theta$

Hence proved.