Question

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

Answer 1

LHS = $tan(\alpha -\beta )=\frac{tan\alpha -tan\beta }{1+tan\alpha tan\beta }$
$\frac{tan\alpha -\frac{nsin\alpha cos\alpha }{1–nsi{n}^2\alpha }}{1+tan\alpha .\frac{nsin\alpha cos\alpha }{1–nsi{n}^2\alpha }}$

$=\frac{tan\alpha –nsi{n}^2\alpha tan\alpha –nsin\alpha cos\alpha }{1–nsi{n}^{2}alpha+ntan\alpha sin\alpha cos\alpha }$

$=\frac{tan\alpha (1–nsi{n}^2\alpha –nco{s}^2\alpha }{tan\alpha (cot\alpha –nsin\alpha cos\alpha +nsin\alpha cos\alpha }$

$=\frac{1–n(si{n}^2\alpha +co{s}^2\alpha )}{cot\alpha }$

$=\frac{1-n}{cot\alpha }$

$=(1-n)tan\alpha$

= RHS