Q62 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

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

  • A
    $\frac{1}{2 \sqrt{2}}$
  • B
    $3 \sqrt{2}$
  • C
    $\frac{1}{\sqrt{2}}$
  • Correct answer
    D
    $\sqrt{2}$

Solution

Given, $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x$ $=\cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$ $\cos 2 x\left[\cos \left(\frac{\pi}{4}-x\right)+\cos \left(\frac{\pi}{4}+x\right)\right]$ $=\sec x \sin 2 x(\cos x-\sin x)$ $\cos 2 x\left(2 \sin \frac{\pi}{4} \sin x\right)=\sec x \sin 2 x(\cos x-\sin x)$ $\frac{2}{\sqrt{2}}\left(\cos ^2 x-\sin ^2 x\right) \sin x=\sec x(2 \sin x \cos x)$ $(\cos x-\sin x)$ $\cos x+\sin x=\sqrt{2}$ $\because \quad \sin x=\cos x=\frac{1}{\sqrt{2}}$ $\sec x=\sqrt{2}$

Hence, the correct answer is option: 'D'.

Relevant Questions