Question

Solve for x:
\[2{{\tan }^{-1}}\left( \sin x \right)={{\tan }^{-1}}\left( 2\sec x \right),x\ne \dfrac{\pi }{2}\]

Answer 1

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


Complete step-by-step answer:
Given that \[2{{\tan }^{-1}}\left( \sin x \right)={{\tan }^{-1}}\left( 2\sec x \right),x\ne \dfrac{\pi }{2}\]
Applying tan on both left hand side and right hand side we will get,
\[\tan \left( 2{{\tan }^{-1}}\left( \sin x \right) \right)=\tan \left( {{\tan }^{-1}}\left( 2\sec x \right) \right)\] . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know that the formula for \[\tan 2\theta \] is given by \[\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }\]
\[\Rightarrow \dfrac{2\tan \left( {{\tan }^{-1}}\left( \sin x \right) \right)}{1-{{\tan }^{2}}\left( {{\tan }^{-1}}\left( \sin x \right) \right)}=2\sec x\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
\[\Rightarrow \dfrac{2\sin x}{1-{{\sin }^{2}}x}=2\sec x\]. . . . . . . . . . . . . . . . . . . . . . . . . (3)
\[\Rightarrow \dfrac{\sin x}{{{\cos }^{2}}x}=\dfrac{1}{\cos x}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . .(4)
\[\Rightarrow \tan x=1\]
The general solution of \[\tan x=k\]is \[n\pi +\alpha \]
\[\Rightarrow x=n\pi +\dfrac{\pi }{4}\]

Note: The basic trigonometric identity is \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]. Note that in the general solution of the equation \[\tan x=k\]which is \[n\pi +\alpha \]and value \[\alpha \]should be in the principal solution that is \[\alpha \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\]. 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.