Question

Show that $y\sin \varphi =x\sin (2\theta +\varphi ),$ then $(x+y)\cot (\theta +\varphi )=(y-x)\cot \theta$.

Answer 1

$y\sin \varphi =x\sin (2\theta +\varphi )$

$\Rightarrow \frac{x}{y}=\frac{\sin \varphi }{\sin (2\theta +\varphi )}$

Applying componendo and dividendo
$\frac{x+y}{x-y}=\frac{\sin \varphi +\sin (2\theta +\varphi )}{\sin \varphi -\sin (2\theta +\varphi )}$

$\frac{x+y}{x-y}=\frac{2\sin \left(\frac{2\theta +2\varphi }{2}\right)\cos \left(\frac{\varphi -2\theta -\varphi }{2}\right)}{2\cos \left(\frac{2\theta +2\varphi }{2}\right)\sin \left(\frac{\varphi -2\theta -\varphi }{2}\right)}=-\frac{\cot \theta }{\cot (\theta +\varphi )}$ ....... $\because \frac{\cos A}{\sin A}=\cot A$

$\therefore (x+y)\left(\cot (\theta +\varphi )\right)=(y-x)\cot \theta$