Question

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

Answer 1

$\tan \beta =\frac{n\sin \alpha \cos \alpha }{1-n{\sin }^2\alpha }=\frac{n\tan \alpha }{{\sec }^2\alpha -n{\tan }^2\alpha }$
$=\frac{n\tan \alpha }{1+{\tan }^2\alpha -n{\tan }^2\alpha }=\frac{n\tan \alpha }{1+(1-n){\tan }^2\alpha }$
$\therefore \tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1-\tan \alpha \tan \beta }$
$=\frac{\tan \alpha -\frac{n\tan \alpha }{1+(1-n){\tan }^2\alpha }}{1-\frac{\tan \alpha n\tan \beta }{1-\tan \alpha \tan \beta }}=(1-n)\tan \alpha$