Question

Solve the inverse trigonometric function for $x$ , $2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( 2\csc x \right)$ .

Answer 1

Hint: For solving this question first we will apply $\tan$ function on both sides of the given equation. After that, we will use the formulas like $\tan \left( {{\tan }^{-1}}x \right)=x$ , $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ and $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta$ to solve further. Then, we will use the result of the equation $\cot x=\cot y$ for writing the final answer for this question.

Complete step-by-step solution -
Given:
We have to find the suitable values of $x$ and we have the following equation:
$2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( 2\csc x \right)$
Now, let ${{\tan }^{-1}}\left( \cos x \right)=\alpha$ and ${{\tan }^{-1}}\left( 2\csc x \right)=\beta$ . And as we know that, $\tan \left( {{\tan }^{-1}}x \right)=x$ where $x\in R$ . Moreover, for any real $x$ $\cos x\in R$ and $\csc x\in R$. Then,
$\begin{aligned} & \alpha ={{\tan }^{-1}}\left( \cos x \right) \\\ & \Rightarrow \tan \alpha =\tan \left( {{\tan }^{-1}}\left( \cos x \right) \right) \\\ & \Rightarrow \tan \alpha =\cos x............................\left( 1 \right) \\\ & \beta ={{\tan }^{-1}}\left( 2\csc x \right) \\\ & \Rightarrow \tan \beta =\tan \left( {{\tan }^{-1}}\left( 2\csc x \right) \right) \\\ & \Rightarrow \tan \beta =2\csc x.........................\left( 2 \right) \\\ \end{aligned}$
Now, before we proceed we should know the following formulas:
$\begin{aligned} & \tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }......................\left( 3 \right) \\\ & 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta ........................\left( 4 \right) \\\ & \csc \theta =\dfrac{1}{\sin \theta }...............................\left( 5 \right) \\\ & \dfrac{\cos \theta }{\sin \theta }=\cot \theta ...............................\left( 6 \right) \\\ & \cot \dfrac{\pi }{4}=1.....................................\left( 7 \right) \\\ \end{aligned}$
Now, we have an equation $2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( 2\csc x \right)$ and as per our assumption, ${{\tan }^{-1}}\left( \cos x \right)=\alpha$ and ${{\tan }^{-1}}\left( 2\csc x \right)=\beta$ . Then,
$\begin{aligned} & 2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( 2\csc x \right) \\\ & \Rightarrow 2\alpha =\beta \\\ \end{aligned}$
Now, we apply $\tan$ function on both sides in the above equation. Then,
$\begin{aligned} & 2\alpha =\beta \\\ & \Rightarrow \tan \left( 2\alpha \right)=\tan \beta \\\ \end{aligned}$
Now, we will use the formula from the equation (3) to write $\tan 2\alpha =\dfrac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }$ in the above equation. Then,
$\begin{aligned} & \tan \left( 2\alpha \right)=\tan \beta \\\ & \Rightarrow \dfrac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }=\tan \beta \\\ \end{aligned}$
Now, we can write $\tan \alpha =\cos x$ from equation (1) and $\tan \beta =2\csc x$ from equation (2) in the above equation. Then,
$\begin{aligned} & \dfrac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }=\tan \beta \\\ & \Rightarrow \dfrac{2\cos x}{1-{{\cos }^{2}}x}=2\csc x \\\ \end{aligned}$
Now, we will use the formula from the equation (4) to write $1-{{\cos }^{2}}x={{\sin }^{2}}x$ and formula from the equation (5) to write $\csc x=\dfrac{1}{\sin x}$ in the above equation. Then,
$\begin{aligned} & \dfrac{2\cos x}{1-{{\cos }^{2}}x}=2\csc x \\\ & \Rightarrow \dfrac{\cos x}{{{\sin }^{2}}x}=\dfrac{1}{\sin x} \\\ & \Rightarrow \dfrac{\cos x}{{{\sin }^{2}}x}-\dfrac{1}{\sin x}=0 \\\ & \Rightarrow \dfrac{1}{\sin x}\left( \dfrac{\cos x}{\sin x}-1 \right)=0 \\\ \end{aligned}$
Now, we will use the formula from the equation (6) to write $\dfrac{\cos x}{\sin x}=\cot x$ and formula from the equation (7) to write $1=\cot \dfrac{\pi }{4}$ in the above equation. Then,
$\begin{aligned} & \dfrac{1}{\sin x}\left( \dfrac{\cos x}{\sin x}-1 \right)=0 \\\ & \Rightarrow \dfrac{1}{\sin x}\left( \cot x-\cot \dfrac{\pi }{4} \right)=0 \\\ \end{aligned}$
Now, from the above result, we conclude that $\cot x=\cot \dfrac{\pi }{4}$ and $\sin x\ne 0$ . Then,
$\cot x=\cot \dfrac{\pi }{4}.............\left( 8 \right)$
Now, before we proceed we should know one important result which we will use here.
If $\cot x=\cot y$ , then the general solution for $x$ in terms of y can be written as,
$x=n\pi +y............\left( 9 \right)$ , where $n$ is any integer.
From (8) we have:
$\begin{aligned} & \cot x=\cot \dfrac{\pi }{4} \\\ & \Rightarrow x=n\pi +\dfrac{\pi }{4} \\\ \end{aligned}$
Now, from the above result we conclude that suitable values of $x$ will be $x=n\pi +\dfrac{\pi }{4}$ , where $n$ is any integer and values of $x$ will be $\dfrac{\pi }{4},\dfrac{5\pi }{4},\dfrac{9\pi }{4}.............n\pi +\dfrac{\pi }{4}$ .
Thus, if $2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( 2\csc x \right)$ then, suitable values of $x$ will be $\dfrac{\pi }{4},\dfrac{5\pi }{4},\dfrac{9\pi }{4}.............n\pi +\dfrac{\pi }{4}$ , where $n$ is any integer.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. Moreover, we should correctly apply every formula without any mathematical error while solving and in the end, when we got equation $\cot x=\cot \dfrac{\pi }{4}$ then, avoid writing $x=\dfrac{\pi }{4}$ directly and solve correctly.