Question

Prove that:
$\frac{(1-2{\sin }^{2}A{)}^2}{{\cos }^{4}A-{\sin }^{4}A}=2{\cos }^{2}A-1$

Answer 1

To prove:- $\frac{{(1-2{\sin }^{2}A)}^2}{{\cos }^{4}A-{\sin }^{4}A}=2{\cos }^{2}A-1$

Proof:-

We know that,

$\cos 2A=2{\cos }^{2}A-1=1-2{\sin }^{2}A={\cos }^{2}A-{\sin }^{2}A$

Now,

Taking L.H.S.-

$\frac{{(1-2{\sin }^{2}A)}^2}{{\cos }^{4}A-{\sin }^{4}A}$

$=\frac{{\cos }^{2}2A}{({\cos }^{2}A-{\sin }^{2}A)({\cos }^{2}A+{\sin }^{2}A)}$

$=\frac{{\cos }^{2}2A}{\cos 2A}$

$=\cos 2A$

$=2{\cos }^{2}A-1$

$=$ R.H.S.

Hence proved