Q61 / 150 · MHT CET 2020 (14 Oct Shift 1) · Mathematics

$\cos \left(\frac{3 \pi}{4}+x\right)-\sin \left(\frac{\pi}{4}-x\right)=$

  • Correct answer
    A
    $-\sqrt{2} \cos x$
  • B
    $-\sqrt{2} \sin x$
  • C
    $\sqrt{2} \cos x$
  • D
    $\sqrt{2} \sin x$

Solution

$\cos \left(\frac{3 \pi}{4}+x\right)-\sin \left(\frac{\pi}{4}-x\right)$ $=\left(\cos \frac{3 \pi}{4} \cos x-\sin \frac{3 \pi}{4} \sin x\right)-\left(\sin \frac{\pi}{4} \cos x-\cos \frac{\pi}{4} \sin x\right)$ $=\frac{-1}{\sqrt{2}} \cos x-\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x$ $=-\sqrt{2} \cos x$

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

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