Question

Find the value of x satisfying the following inverse trigonometric equation
$2{{\tan }^{-1}}\cos x={{\tan }^{-1}}\left( 2\csc x \right)$

Answer 1

Hint: Take tan on both sides of the equation and use the fact that $\tan \left( {{\tan }^{-1}}x \right)=x$ and $\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$. Use $1-{{\cos }^{2}}x={{\sin }^{2}}x$ and $\csc x=\dfrac{1}{\sin x}$ and hence prove that $\tan \left( 2{{\tan }^{-1}}\cos x \right)=\dfrac{\cos x}{{{\sin }^{2}}x}$ and $\tan \left( {{\tan }^{-1}}\left( 2\csc x \right) \right)=\dfrac{2}{\sin x}$. Hence prove that the given equation is equivalent to $\cos x=\sin x$. Divide both sides of the equation by cosx and use the fact that if $\tan x=\tan y$ then $x=n\pi +y,n\in \mathbb{Z}$. Remove the extraneous roots and hence find the solution of the given equation. Verify by substituting in the given equation.

Complete step-by-step solution -
We have $2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( 2\csc x \right)$
Taking tangents on both sides, we get
$\tan \left( 2{{\tan }^{-1}}\cos x \right)=\tan \left( {{\tan }^{-1}}\left( 2\csc x \right) \right)\text{ }\left( i \right)$
Simplifying $\tan \left( 2{{\tan }^{-1}}\cos x \right)$
We know that $\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$
Hence, we have
$\tan \left( 2{{\tan }^{-1}}\cos x \right)=\dfrac{2\tan \left( {{\tan }^{-1}}\cos x \right)}{1-{{\tan }^{2}}\left( {{\tan }^{-1}}\cos x \right)}$
We know that $\tan \left( {{\tan }^{-1}}x \right)=x\forall x\in \mathbb{R}$
Using the above identity, we get
$\tan \left( 2{{\tan }^{-1}}\cos x \right)=\dfrac{2\cos x}{1-{{\cos }^{2}}x}$
We know that $1-{{\cos }^{2}}x={{\sin }^{2}}x$
Hence, we have
$\tan \left( 2{{\tan }^{-1}}\cos x \right)=\dfrac{2\cos x}{{{\sin }^{2}}x}$
Simplifying $\tan \left( {{\tan }^{-1}}2\csc x \right)$
We know that $\tan \left( {{\tan }^{-1}}x \right)=x\forall x\in \mathbb{R}$
Hence, we have
$\tan \left( {{\tan }^{-1}}2\csc x \right)=2\csc x$
We know that $\csc x=\dfrac{1}{\sin x}$
Using the above identity, we get
$\tan \left( {{\tan }^{-1}}\left( 2\csc x \right) \right)=\dfrac{2}{\sin x}$
Hence the equation(i) becomes
$\dfrac{\cos x}{{{\sin }^{2}}x}=\dfrac{1}{\sin x}$
Multiplying both sides by ${{\sin }^{2}}x$, we get
$\cos x=\sin x$
Dividing both sides by cosx, we get
$\dfrac{\sin x}{\cos x}=1$
We know that $\dfrac{\sin x}{\cos x}=\tan x$
Hence, we have
$\tan x=1$
We know that $\tan \left( \dfrac{\pi }{4} \right)=1$
Hence, we have
$\tan x=\tan \left( \dfrac{\pi }{4} \right)$
We know that if tanx=tany, then $x=n\pi +y,n\in \mathbb{Z}$
Hence, we have
$x=n\pi +\dfrac{\pi }{4},n\in \mathbb{Z}$ which is the required solution of the given equation.

Note: Verification:
We have if n is odd
$\cos \left( x \right)=\dfrac{-1}{\sqrt{2}}$ and $\csc x=-\sqrt{2}$
Hence, we have
${{\tan }^{-1}}\left( 2\csc x \right)=-{{\tan }^{-1}}2\sqrt{2}=-{{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{\sqrt{2}}}{1-{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}} \right)=-2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}}$ and $2{{\tan }^{-1}}\cos x=-2{{\tan }^{-1}}\dfrac{1}{\sqrt{2}}$
Hence, we have
${{\tan }^{-1}}\left( 2\csc x \right)=2{{\tan }^{-1}}\left( \cos x \right)$
Similarly when n is even ${{\tan }^{-1}}\left( 2\csc x \right)=2{{\tan }^{-1}}\left( \cos x \right)$
Hence our answer is verified to be correct.