Question

If $\sin \theta =\frac{x-y}{x+y}$ then show that $\tan (\frac{\pi }{4}-\frac{\theta }{2})=\pm \sqrt{\frac{y}{x}}$.

Answer 1

We have,

$\sin \theta =\frac{x-y}{x+y}$

${\cos }^2\theta =1-{\sin }^2\theta$

${\cos }^2\theta =1-{\left(\frac{x-y}{x+y}\right)}^2$

${\cos }^2\theta =\frac{{(x+y)}^2-{(x-y)}^2}{{(x+y)}^2}$

$\cos \theta =\pm \frac{2\sqrt{xy}}{x+y}$

Taking L.H.S.

$=\tan (\frac{\pi }{4}-\frac{\theta }{2})$

$=\frac{\tan \frac{\pi }{4}-\tan \frac{\theta }{2}}{1+\tan \frac{\pi }{4}\tan \frac{\theta }{2}}$

$=\frac{1-\tan \frac{\theta }{2}}{1+\tan \frac{\theta }{2}}$

$=\frac{1-\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}}{1+\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}}$

$=\frac{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}$

$=\frac{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}\times \frac{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}$

$=\frac{1-\sin \frac{2\theta }{2}}{\cos \frac{2\theta }{2}}$

$=\frac{1-\sin \theta }{\cos \theta }$ ……. (1)

On putting the value of $\sin \theta$ and $\cos \theta$ in equation (1), we get

$\tan (\frac{\pi }{4}-\frac{\theta }{2})=\frac{1-\sin \theta }{\cos \theta }$

$=\frac{1-\frac{x-y}{x+y}}{\pm \frac{2\sqrt{xy}}{x+y}}$

$=\pm \frac{x+y-(x-y)}{2\sqrt{xy}}$

$=\pm \frac{2y}{2\sqrt{xy}}$

$=\pm \sqrt{\frac{y}{x}}$

$\tan (\frac{\pi }{4}-\frac{\theta }{2})=\pm \sqrt{\frac{y}{x}}$

Hence, proved.