Question

If $\sin \theta =\frac{x-y}{x+y}$ show that tan $\left(\frac{\pi }{4}-\frac{0}{2}\right)=+\sqrt{\frac{y}{x}}$

Answer 1

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

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

$=(\tan \left(\frac{\pi }{4}\right)-\tan \theta (2)/1+\tan \frac{\pi }{4}\ast \tan \theta /2)$

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

$=(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})/(\cos \frac{\theta }{2}+\sin \frac{\pi }{2})$

$=\frac{{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}^2}{{\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}}$

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

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

$1-\sin \theta =\frac{n+y-(x-y)}{x+y}=\frac{2y}{x+y}$

${\cos }^2\theta =1-{\sin }^2=\frac{(x+y{)}^2-(x-y{)}^2}{(x+y{)}^2}=\frac{4xy}{(x+y{)}^2}$

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

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