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If $\sin ^{4} x-\cos ^{4} x=p$, then which one of the following is correct?
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Verified Answer
The correct answer is:
$|\mathrm{p}| \leq 1$
Consider $\sin ^{4} x-\cos ^{4} x=p$
$\Rightarrow\left(\sin ^{2} x\right)^{2}-\left(\cos ^{2} x\right)^{2}=p$
$\Rightarrow\left(\sin ^{2} x-\cos ^{2} x\right)\left(\sin ^{2} x+\cos ^{2} x\right)=p$
$\Rightarrow \sin ^{2} x-\cos ^{2} x=p\left(\because \sin ^{2} x+\cos ^{2} x=1\right)$
$\Rightarrow-\cos 2 x=p \quad\left(\because \cos ^{2} x-\sin ^{2} x=\cos 2 x\right)$
$\Rightarrow \cos 2 x=-p$
$\therefore \quad|p| \leq 1$
$\Rightarrow\left(\sin ^{2} x\right)^{2}-\left(\cos ^{2} x\right)^{2}=p$
$\Rightarrow\left(\sin ^{2} x-\cos ^{2} x\right)\left(\sin ^{2} x+\cos ^{2} x\right)=p$
$\Rightarrow \sin ^{2} x-\cos ^{2} x=p\left(\because \sin ^{2} x+\cos ^{2} x=1\right)$
$\Rightarrow-\cos 2 x=p \quad\left(\because \cos ^{2} x-\sin ^{2} x=\cos 2 x\right)$
$\Rightarrow \cos 2 x=-p$
$\therefore \quad|p| \leq 1$
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