Q95 / 150 · MHT CET 2021 (22 Sep Shift 2) · Mathematics

If $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$, then the value of $x$ is

  • A
    $\frac{\pi^c}{6}$
  • Correct answer
    B
    $\frac{\pi^c}{4}$
  • C
    $\frac{\pi^c}{3}$
  • D
    $\frac{\pi^\epsilon}{12}$

Solution

We have $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$ $$ \begin{aligned} & \therefore \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}\left(\frac{2}{\sin x}\right) \\ & \therefore \frac{2 \cos x}{\sin ^2 x}=\frac{2}{\sin x} \Rightarrow \sin x \cos x=\sin ^2 x \\ & \therefore \sin x(\sin x-\cos x)=0 \Rightarrow \sin x=0 \text { or } \tan x=1 \\ & \therefore x=0 \text { or } x=\frac{\pi}{4} \end{aligned} $$

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

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