Question

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

Answer 1

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

$\Rightarrow \frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=\frac{x-y}{x+y}$

$\Rightarrow 2\tan \frac{\theta }{2}x+2\tan \frac{\theta }{2}y=x-y+x{\tan }^2\frac{\theta }{2}-y{\tan }^2\frac{\theta }{2}$

$\Rightarrow 2\tan \frac{\theta }{2}x+2\tan \frac{\theta }{2}y-x+y-x{\tan }^2\frac{\theta }{2}+y{\tan }^2\frac{\theta }{2}=0$

$\Rightarrow 2\tan \frac{\theta }{2}x-x-x{\tan }^2\frac{\theta }{2}=-2\tan \frac{\theta }{2}y-y-y{\tan }^2\frac{\theta }{2}$

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

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

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

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

$\Rightarrow \frac{(\tan \frac{\theta }{2}-1)}{(\tan \frac{\theta }{2}+1)}=\pm \sqrt{\frac{y}{x}}$

$\Rightarrow \frac{(\tan \frac{\theta }{2}-\tan \frac{\pi }{4})}{(\tan \frac{\theta }{2}\tan \frac{\pi }{4}+1)}=\pm \sqrt{\frac{y}{x}}$

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

Hence proved.