Question

If $\theta +\varphi =\alpha$ and $\tan \theta =k\tan \varphi$, then prove that $\sin \left(\theta -\varphi \right)=\frac{k-1}{k+1}\sin \alpha$.

Answer 1

$\theta +\varphi =\alpha \tan \theta =\tan \alpha$

$\frac{\tan \theta }{\tan \varphi }=\frac{k}{1}$

Applying componendo and dividendo

$\frac{\tan \theta -\tan \varphi }{\tan \theta +\tan \varphi }=\frac{k-1}{k+1}$

$\frac{\frac{\sin \theta }{\cos \theta }-\frac{\sin \varphi }{\cos \varphi }}{\frac{\sin \theta }{\cos \theta }+\frac{\sin \varphi }{\cos \varphi }}=\frac{k-1}{k+1}$

$\frac{\sin \left(\theta -\varphi \right)}{\sin \left(\theta +\varphi \right)}=\frac{k-1}{k+1}$

$\frac{\sin \left(\theta -\varphi \right)}{\sin \alpha }=\frac{k-1}{k+1}$

$\sin \left(\theta -\varphi \right)=\left(\frac{k-1}{k+1}\right)\sin \alpha$