Question

If ${\tan }^2\alpha ={\cos }^2\beta -{\sin }^2\beta$.then prove that ${\cos }^2\alpha -{\sin }^2\alpha ={\tan }^2\beta$

Answer 1

Let

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

or, $\frac{{\sin }^2\alpha }{{\cos }^2\alpha }={\cos }^2\beta -{\sin }^2\beta$

or, $\frac{{\cos }^2\alpha }{{\sin }^2\alpha }=\frac{1}{{\cos }^2\beta -{\sin }^2\beta }$

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

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

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