Q71 / 150 · MHT CET 2022 (07 Aug Shift 2) · Mathematics

The angle between the curves $y=\sin x$ and $y=\cos x 0<x<\frac{\pi}{2}$, is

  • A
    $\tan ^{-1}(\sqrt{2})$
  • B
    $\tan ^{-1}(3 \sqrt{2})$
  • Correct answer
    C
    $\tan ^{-1}(2 \sqrt{2})$
  • D
    $\tan ^{-1}(3 \sqrt{3})$

Solution

$y=\sin x$ and $y=\cos x$ intersects at $x=\frac{\pi}{4}$ $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\text {for first curve at } x=\frac{\pi}{4}}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$ $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\text {for second curve at } x=\frac{\pi}{4}}=-\sin \frac{\pi}{4}=-\frac{1}{\sqrt{2}}$ Required angle $=\theta=\tan ^{-1}\left\{\frac{\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)}{1+\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)}\right\}=\tan ^{-1}\left(\frac{\sqrt{2}}{\frac{1}{2}}\right)$ $=\tan ^{-1}(2 \sqrt{2})$

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

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