Question

Solve the inverse trigonometric equation for x, $2{{\tan }^{-1}}\left( \sin x \right)={{\tan }^{-1}}\left( 2\sec x \right),x\ne \dfrac{\pi }{2}$.

Answer 1

Hint: Use the identity given by: $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ to simplify the term in L.H.S. Once the coefficient of tan inverse becomes the same on both the sides, remove the tan inverse function from both the sides and equate their arguments. Solve the obtained trigonometric equation and find the general solution of the equation. Use the formula for the general solution: if $\tan a=\tan b$, then $a=n\pi +b$, where ‘n’ is any integer.

Complete step-by-step solution -
We have been provided with the equation: $2{{\tan }^{-1}}\left( \sin x \right)={{\tan }^{-1}}\left( 2\sec x \right),x\ne \dfrac{\pi }{2}$.
Here, we have been given the condition: $x\ne \dfrac{\pi }{2}$, because tangent function and secant function is undefined at $x=\dfrac{\pi }{2}$. So, the solution cannot be $x=\dfrac{\pi }{2}$.
Now let us come to the question.
$2{{\tan }^{-1}}\left( \sin x \right)={{\tan }^{-1}}\left( 2\sec x \right)$
Applying the formula: $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$, we have,
${{\tan }^{-1}}\left( \dfrac{2\sin x}{1-{{\sin }^{2}}x} \right)={{\tan }^{-1}}\left( 2\sec x \right)$
Using the identity: $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $, we get,
${{\tan }^{-1}}\left( \dfrac{2\sin x}{{{\cos }^{2}}x} \right)={{\tan }^{-1}}\left( 2\sec x \right)$
Now, removing tan inverse function from both sides, we get,
$\left( \dfrac{2\sin x}{{{\cos }^{2}}x} \right)=\left( 2\sec x \right)$
We know that, $\sec x=\dfrac{1}{\cos x}$, therefore, the expression becomes:
$\left( \dfrac{2\sin x}{{{\cos }^{2}}x} \right)=\left( \dfrac{2}{\cos x} \right)$
Cancelling the common terms, we get,
$\begin{align}
  & \dfrac{\sin x}{\cos x}=1 \\
 & \Rightarrow \tan x=1 \\
 & \Rightarrow \tan x=\tan \dfrac{\pi }{4} \\
\end{align}$
Using the formula for general solution given by: if $\tan a=\tan b$, then $a=n\pi +b$, we get,
$x=n\pi +\dfrac{\pi }{4}$, where ‘n’ is any integer.

Note: One may note that, we cannot directly remove tan inverse function from both sides at the very beginning because the coefficient of this inverse function is not equal on both sides. Therefore, first we made the coefficients of the given tan inverse function equal on both sides and then removed it from both the sides. You may note that we have found the general solution of the given equation because there are no limits provided to us to find any particular solution or principal solution. You may find any particular solution by substituting the integral values of ‘n’.