Question

$co{s}^2\alpha -si{n}^2\alpha =ta{n}^2\beta .$ then show that $ta{n}^2\alpha =co{s}^2\beta -si{n}^2\beta .$

Answer 1

${\cos }^2\alpha -{\sin }^2\alpha ={\tan }^2\beta$

$\Rightarrow \cos 2\alpha ={\tan }^2\beta$

$\Rightarrow \frac{1}{\cos 2\alpha }=\frac{1}{{\tan }^2\beta }$

Applying compound & Dividendo

$\frac{1-\cos 2\alpha }{1+\cos 2\alpha }=\frac{1-{\tan }^2\beta }{1+{\tan }^2\beta }$

$\frac{2{\sin }^2\alpha }{2{\cos }^2\alpha }=\cos 2\beta$ $[1+{\cos }^2\alpha =2{\cos }^2\alpha ,1-\cos 2\alpha =2{\sin }^2\alpha ,\cos 2\beta =\frac{1-{\tan }^2\beta }{1+{\tan }^2\beta }]$

${\tan }^2\alpha =\cos 2\beta$

$\Rightarrow {\tan }^2\alpha ={\cos }^2\beta -{\sin }^2\beta$.