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Question: \(2\tan^{- 1}(\cos x) = \tan^{- 1}(\text{cose}\text{c}^{2}x),\) then x =...

2tan⁡−1(cos⁡x)=tan⁡−1(cosec2x),2\tan^{- 1}(\cos x) = \tan^{- 1}(\text{cose}\text{c}^{2}x), then x =

A

π2\frac{\pi}{2}

B

π\pi

C

π6\frac{\pi}{6}

D

π3\frac{\pi}{3}

Answer

π3\frac{\pi}{3}

Explanation

Solution

2tan⁡−1(cos⁡x)=tan⁡−1(cos⁡ec2x)\mathbf{2}\mathbf{\tan}^{\mathbf{- 1}}\mathbf{(}\mathbf{\cos}\mathbf{x}\mathbf{)}\mathbf{=}\mathbf{\tan}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{\cos}\mathbf{e}\mathbf{c}^{\mathbf{2}}\mathbf{x)}

⇒ tan⁡−1(2cos⁡x1−cos⁡2x)=tan⁡−1(1sin⁡2x)⇒2cos⁡xsin⁡2x=1sin⁡2x\mathbf{\tan}^{\mathbf{-}\mathbf{1}}\left( \frac{\mathbf{2}\mathbf{\cos}\mathbf{x}}{\mathbf{1}\mathbf{-}\mathbf{\cos}^{\mathbf{2}}\mathbf{x}} \right)\mathbf{=}\mathbf{\tan}^{\mathbf{-}\mathbf{1}}\left( \frac{\mathbf{1}}{\mathbf{\sin}^{\mathbf{2}}\mathbf{x}} \right)\mathbf{\Rightarrow}\frac{\mathbf{2}\mathbf{\cos}\mathbf{x}}{\mathbf{\sin}^{\mathbf{2}}\mathbf{x}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\sin}^{\mathbf{2}}\mathbf{x}}

⇒ 2cos⁡x=1⇒x=π32\cos x = 1 \Rightarrow x = \frac{\pi}{3}