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If $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x$ $=\cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$, then a possible value of sec $x$ is
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The correct answer is:
$\sqrt{2}$
Given,
$\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x$ $=\cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$
$\cos 2 x\left[\cos \left(\frac{\pi}{4}-x\right)+\cos \left(\frac{\pi}{4}+x\right)\right]$ $=\sec x \sin 2 x(\cos x-\sin x)$
$\cos 2 x\left(2 \sin \frac{\pi}{4} \sin x\right)=\sec x \sin 2 x(\cos x-\sin x)$
$\frac{2}{\sqrt{2}}\left(\cos ^2 x-\sin ^2 x\right) \sin x=\sec x(2 \sin x \cos x)$ $(\cos x-\sin x)$
$\cos x+\sin x=\sqrt{2}$
$\because \quad \sin x=\cos x=\frac{1}{\sqrt{2}}$
$\sec x=\sqrt{2}$
$\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x$ $=\cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$
$\cos 2 x\left[\cos \left(\frac{\pi}{4}-x\right)+\cos \left(\frac{\pi}{4}+x\right)\right]$ $=\sec x \sin 2 x(\cos x-\sin x)$
$\cos 2 x\left(2 \sin \frac{\pi}{4} \sin x\right)=\sec x \sin 2 x(\cos x-\sin x)$
$\frac{2}{\sqrt{2}}\left(\cos ^2 x-\sin ^2 x\right) \sin x=\sec x(2 \sin x \cos x)$ $(\cos x-\sin x)$
$\cos x+\sin x=\sqrt{2}$
$\because \quad \sin x=\cos x=\frac{1}{\sqrt{2}}$
$\sec x=\sqrt{2}$
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