Question

Prove the following identities:

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

Answer 1

L.H.S. ${\displaystyle \frac{{(1-2{\sin }^{2}A)}^2}{{\cos }^{4}A-{\sin }^{4}A}}$

= ${\displaystyle \frac{{(1-2{\sin }^{2}A)}^2}{{\left({\cos }^{2}A\right)}^2-{\left({\sin }^{2}A\right)}^2}}$

= ${\displaystyle \frac{{(1-2{\sin }^{2}A)}^2}{({\cos }^{2}A+{\sin }^{2}A)({\cos }^{2}A-{\sin }^{2}A)}}$  

= ${\displaystyle \frac{{(1-2{\sin }^{2}A)}^2}{\left(1\right)(1-{\sin }^{2}A-{\sin }^{2}A)}}$   ...[∵ cos² A = 1 – sin² A]

= ${\displaystyle \frac{{(1-2{\sin }^{2}A)}^2}{(1-2{\sin }^{2}A)}}$

= 1 – 2 sin² A

= 1 – 2 (1 – cos² A) 

= 1 – 2 + 2 cos² A

= 2 cos² A – 1 = R.H.S.

Hence the result is proved.