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Question: Solve for x: \[2{{\tan }^{-1}}\left( \sin x \right)={{\tan }^{-1}}\left( 2\sec x \right),x\ne \dfr...

Solve for x:
2tan⁡−1(sin⁡x)=tan⁡−1(2sec⁡x),x≠π22{{\tan }^{-1}}\left( \sin x \right)={{\tan }^{-1}}\left( 2\sec x \right),x\ne \dfrac{\pi }{2}

Explanation

Solution

Hint: Apply the tan on both the left hand side and right hand side. Now apply the formula for tan⁡2θ\tan 2\theta on the left hand side. Now use inverse trigonometric functions that is tan⁡(tan⁡−1θ)=θ\tan \left( {{\tan }^{-1}}\theta \right)=\theta and use basic trigonometric identities to get tan⁡x\tan xand write the general solution of the equation tan⁡x=k\tan x=k.

Complete step-by-step answer:
Given that 2tan⁡−1(sin⁡x)=tan⁡−1(2sec⁡x),x≠π22{{\tan }^{-1}}\left( \sin x \right)={{\tan }^{-1}}\left( 2\sec x \right),x\ne \dfrac{\pi }{2}
Applying tan on both left hand side and right hand side we will get,
tan⁡(2tan⁡−1(sin⁡x))=tan⁡(tan⁡−1(2sec⁡x))\tan \left( 2{{\tan }^{-1}}\left( \sin x \right) \right)=\tan \left( {{\tan }^{-1}}\left( 2\sec x \right) \right) . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know that the formula for tan⁡2θ\tan 2\theta is given by tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }
⇒2tan⁡(tan⁡−1(sin⁡x))1−tan⁡2(tan⁡−1(sin⁡x))=2sec⁡x\Rightarrow \dfrac{2\tan \left( {{\tan }^{-1}}\left( \sin x \right) \right)}{1-{{\tan }^{2}}\left( {{\tan }^{-1}}\left( \sin x \right) \right)}=2\sec x. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
⇒2sin⁡x1−sin⁡2x=2sec⁡x\Rightarrow \dfrac{2\sin x}{1-{{\sin }^{2}}x}=2\sec x. . . . . . . . . . . . . . . . . . . . . . . . . (3)
⇒sin⁡xcos⁡2x=1cos⁡x\Rightarrow \dfrac{\sin x}{{{\cos }^{2}}x}=\dfrac{1}{\cos x}. . . . . . . . . . . . . . . . . . . . . . . . . . . . .(4)
⇒tan⁡x=1\Rightarrow \tan x=1
The general solution of tan⁡x=k\tan x=kis nπ+αn\pi +\alpha
⇒x=nπ+π4\Rightarrow x=n\pi +\dfrac{\pi }{4}

Note: The basic trigonometric identity is sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1. Note that in the general solution of the equation tan⁡x=k\tan x=kwhich is nπ+αn\pi +\alpha and value α\alpha should be in the principal solution that is α∈(−π2,π2)\alpha \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right). Since the trigonometric functions are periodic functions, these functions are not bijections in their natural domains. Therefore the inverse function does not exist. By identifying the proper domains they are bijections and so an inverse function exists.