Question
Prove that $\sin^4A – \cos^4A = 1 – 2 \cos^2A$
✓
Answer
$ \text { L.H.S }=\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) \ldots \ldots\left[\because a^2-b^2=(a+b)(a-b)\right]$
$ =(1)\left(\sin ^2 A-\cos ^2 A\right) \ldots \ldots\left[\because \sin ^2 A+\cos ^2 A=1\right]$
$ =\sin ^2 A-\cos ^2 A$
$=\left(1-\cos ^2 A\right)-\cos ^2 A \ldots \ldots\left[\begin{array}{l}\left.\sin ^2 A+\cos ^2 A=1\right] \\ \left.\therefore 1-\cos ^2=\sin ^2 A\right]\end{array}\right.$
$ =1-2 \cos ^2 A$
$ =\text { R.H.S }$
$ \therefore \sin ^4 A-\cos ^4 A=1-2 \cos ^2 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) \ldots \ldots\left[\because a^2-b^2=(a+b)(a-b)\right]$
$ =(1)\left(\sin ^2 A-\cos ^2 A\right) \ldots \ldots\left[\because \sin ^2 A+\cos ^2 A=1\right]$
$ =\sin ^2 A-\cos ^2 A$
$=\left(1-\cos ^2 A\right)-\cos ^2 A \ldots \ldots\left[\begin{array}{l}\left.\sin ^2 A+\cos ^2 A=1\right] \\ \left.\therefore 1-\cos ^2=\sin ^2 A\right]\end{array}\right.$
$ =1-2 \cos ^2 A$
$ =\text { R.H.S }$
$ \therefore \sin ^4 A-\cos ^4 A=1-2 \cos ^2 A$
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