Question

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

Answer 1

Given,

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

or, $\cos 2\theta ={\tan }^2\alpha$ [ Since, $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$]

or, $\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\frac{{\tan }^2\alpha }{1}$ [ As $\cos 2\theta =\frac{1+{\tan }^2\theta }{1+{\tan }^2\theta }$]

or, $\frac{2}{2{\tan }^2\theta }=\frac{1+{\tan }^2\alpha }{1-{\tan }^2\alpha }$[ By componendo-dividendo]

or, ${\tan }^2\theta =\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$

or, ${\tan }^2\theta =\cos 2\alpha$

or, ${\tan }^2\theta ={\cos }^2\alpha -{\sin }^2\alpha$.