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Miscellaneous Exercise · Q11

Q.Find the value of the following: 2tan⁡−1(cos⁡x)=tan⁡−1(2cosec⁡x)2\tan^{-1} (\cos x) = \tan^{-1} (2 \operatorname{cosec} x)

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Appeared in past exams:MHT-CET 2025· Set pcm-2025-04-20-E· 2mexactMHT-CET 2021· Set pcm-2021-09-22-E· 2mexact
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Applying the double-angle identity for tan⁡−1\tan^{-1} reduces the equation to cos⁡x=sin⁡x\cos x = \sin x, giving x=π4x = \dfrac{\pi}{4} (general solution x=nπ+π4x = n\pi + \tfrac{\pi}{4}).

We solve

2tan⁡−1(cos⁡x)=tan⁡−1(2cosec⁡x).2\tan^{-1}(\cos x) = \tan^{-1}(2\operatorname{cosec} x).

Step 1 — Use the double-angle identity

With 2tan⁡−1u=tan⁡−1 ⁣(2u1−u2)2\tan^{-1} u = \tan^{-1}\!\left(\dfrac{2u}{1-u^2}\right) and u=cos⁡xu = \cos x, the left side becomes

2tan⁡−1(cos⁡x)=tan⁡−1 ⁣(2cos⁡x1−cos⁡2x)=tan⁡−1 ⁣(2cos⁡xsin⁡2x).2\tan^{-1}(\cos x) = \tan^{-1}\!\left(\frac{2\cos x}{1-\cos^2 x}\right) = \tan^{-1}\!\left(\frac{2\cos x}{\sin^2 x}\right).

The equation is now

tan⁡−1 ⁣(2cos⁡xsin⁡2x)=tan⁡−1(2cosec⁡x).\tan^{-1}\!\left(\frac{2\cos x}{\sin^2 x}\right) = \tan^{-1}(2\operatorname{cosec} x).

Step 2 — Equate the arguments

Since tan⁡−1\tan^{-1} is one-to-one,

2cos⁡xsin⁡2x=2cosec⁡x=2sin⁡x.\frac{2\cos x}{\sin^2 x} = 2\operatorname{cosec} x = \frac{2}{\sin x}.

Step 3 — Simplify

Multiply both sides by sin⁡2x2\dfrac{\sin^2 x}{2} (valid since sin⁡x≠0\sin x \ne 0):

cos⁡x=sin⁡x⟹tan⁡x=1.\cos x = \sin x \quad\Longrightarrow\quad \tan x = 1.

Step 4 — Solve …

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