Question

Given $\tan \theta =\frac{n\sin \alpha }{m+n\cos \alpha }$ then show that $\tan \left(\alpha -\theta \right)=\frac{m\sin \alpha }{m\cos \alpha +n}$

Answer 1

$\tan \theta =\frac{n\sin \alpha }{m+n\cos \alpha }$

$\tan \left(\alpha -\theta \right)=\frac{\tan \alpha -\tan \theta }{1+\tan \alpha \tan \theta }$

$=\frac{\tan \alpha -\frac{n\sin \alpha }{m+n\cos \alpha }}{1+\tan \alpha .\frac{n\sin \alpha }{m+n\cos \alpha }}$

$=\frac{\left(m+n\cos \alpha \right)\tan \alpha -n\sin \alpha }{m+n\cos \alpha +n\sin \alpha \tan \alpha }$

$=\frac{m\tan \alpha +n\sin \alpha -n\sin \alpha }{m+n\left(\cos \alpha +\frac{{\sin }^2\alpha }{\cos \alpha }\right)}$

$=\frac{m\tan \alpha }{m+\frac{n}{\cos \alpha }}=\frac{m\sin \alpha }{m\cos \alpha +n}$