Question
Prove.
$\sin ^4 A-\cos ^4 A=2 \sin ^2 A-1$
$\sin ^4 A-\cos ^4 A=2 \sin ^2 A-1$
✓
Answer
$\text { LHS }=\sin ^4 A-\cos ^4 A $
$ =\left(\sin ^2 A\right)^2-\left(\cos ^2 A\right)^2$
$ =\left(\sin ^2 A+\cos ^2 A\right)\left(\sin ^2 A-\cos ^2 A\right)$
$ =\sin ^2 A-\cos ^2 A $
$=\sin ^2 A-\left(1-\sin ^2 A\right) $
$=2 \sin ^2 A-1 \text { RHS }$
$ =\left(\sin ^2 A\right)^2-\left(\cos ^2 A\right)^2$
$ =\left(\sin ^2 A+\cos ^2 A\right)\left(\sin ^2 A-\cos ^2 A\right)$
$ =\sin ^2 A-\cos ^2 A $
$=\sin ^2 A-\left(1-\sin ^2 A\right) $
$=2 \sin ^2 A-1 \text { RHS }$
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