Question

If y sin ϕ = x sin (2θ + ϕ), prove that (x + y) cot (θ + ϕ) = (y − x) cot θ.

Answer 1

Given:
y sin ϕ = x sin (2θ + ϕ)

$$\begin{gathered} \Rightarrow \frac{y}{x}=\frac{\sin \left(2\theta +\varphi \right)}{\sin \varphi } \\ \mathrm{Applying}\mathrm{componendo}\mathrm{and}\mathrm{dividendo}: \\ \Rightarrow \frac{y-x}{y+x}=\frac{\sin \left(2\theta +\varphi \right)-\sin \varphi }{\sin \left(2\theta +\varphi \right)+\sin \varphi } \\ \Rightarrow \frac{y-x}{y+x}=\frac{2\sin \left({\displaystyle \frac{2\theta +\varphi -\varphi }{2}}\right)\cos \left({\displaystyle \frac{2\theta +\varphi +\varphi }{2}}\right)}{2\sin \left({\displaystyle \frac{2\theta +\varphi +\varphi }{2}}\right)\cos \left({\displaystyle \frac{2\theta +\varphi -\varphi }{2}}\right)} \\ \Rightarrow \frac{y-x}{y+x}=\frac{2\sin \theta \cos \left(\theta +\varphi \right)}{2\sin \left(\theta +\varphi \right)\cos \theta } \\ \Rightarrow \frac{y-x}{y+x}=\frac{\sin \theta \cos \left(\theta +\varphi \right)}{\sin \left(\theta +\varphi \right)\cos \theta } \\ \Rightarrow \frac{y-x}{y+x}=\frac{\cot \left(\theta +\varphi \right)}{\cot \theta } \\ \Rightarrow \left(y-x\right)cot\theta =\left(y+x\right)cot\left(\theta +\varphi \right) \end{gathered}$$