Question

If $\tan \beta =\frac{n\sin \alpha \cos \alpha }{1-n{\sin }^2\alpha }$, show that $\tan (\alpha -\beta )=(1-n)\tan \alpha$.

Answer 1

$\tan \beta =\frac{n\sin \alpha \cos \alpha }{1-n\sin {}^2\alpha }\Rightarrow \tan (\alpha -\beta )=(1-n)\tan \alpha LHS=\tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }=\frac{\tan \alpha -\frac{n\sin \alpha \cos \alpha }{1-n\sin {}^2\alpha }}{1+\frac{\tan \alpha n\sin \alpha \cos \alpha }{1-n\sin {}^2\alpha }}=\frac{\tan \alpha -\frac{n\sin \alpha \cos \alpha }{1-n\sin {}^2\alpha }}{1+\frac{n\sin {}^2\alpha }{1-n\sin {}^2\alpha }}=\frac{\tan \alpha (1-n\sin {}^2\alpha )-n\sin \alpha \cos \alpha }{1-n\sin {}^2\alpha +n\sin {}^2\alpha }=\tan \alpha -n\tan \alpha \sin {}^2\alpha -n\sin \alpha \cos \alpha =\tan \alpha -n\tan \alpha (\sin {}^2\alpha +\cos {}^2\alpha )=\tan (1-n)=RHS$