Question

If $ysin\varphi =xsin(2\theta +\varphi )$, prove that $(x+y)cot(\theta +\varphi )=(y-x)cot\theta$

Answer 1

We have,
$ysin\varphi =xsin(2\theta +\varphi )\dots \left(i\right)\Rightarrow \frac{sin\varphi }{sin(2\theta +\varphi )}=\frac{x}{y}$
Now,
$\frac{sin\varphi }{sin(2\theta +\varphi )}=\frac{x}{y}\Rightarrow \frac{sin\varphi }{sin(2\theta +\varphi )}+1=\frac{x}{y}+1\Rightarrow \frac{sin\varphi +sin(2\theta +\varphi )}{sin(2\theta +\varphi )}=\frac{x+y}{y}\dots \left(ii\right)$
Again,
$\frac{sin\varphi }{sin(2\theta +\varphi )}=\frac{x}{y}$
[By equation(i)]
$\Rightarrow \frac{sin\varphi }{sin(2\theta +\varphi )}-1=\frac{x}{y}-1\Rightarrow \frac{sin\varphi -sin(2\theta +\varphi )}{sin\varphi -sin(2\theta +\varphi )}=\frac{x+y}{x-y}\Rightarrow \frac{2sin\left(\frac{\varphi +2\theta +\varphi }{2}\right)cos\left(\frac{\varphi -2\theta -\varphi }{2}\right)}{2sin\left(\frac{\varphi -2\theta -\varphi }{2}\right)cos\left(\frac{\varphi +2\theta +\varphi }{2}\right)}=\frac{x+y}{x-y}\Rightarrow \frac{sin(\theta +\varphi )cos(\theta -\varphi )}{sin(-\theta )cos(\theta +\varphi )}==\frac{x+y}{x-y}\Rightarrow \frac{sin(\theta +\varphi )cos\left(\theta \right)}{cos(\theta +\varphi )[-sin(\theta \left)\right]}=\frac{x+y}{x-y}\Rightarrow \frac{-cot\left(\theta \right)}{cot(\theta +\varphi )}==\frac{x+y}{x-y}\Rightarrow -(x-y)cot\theta =(x+y)cot(\theta +\varphi )\Rightarrow (y-x)cot\theta =(x+y)cot(\theta +\varphi )\Rightarrow (x+y)cot(\theta +\varphi )=(y-x)cot\theta$