Question

# If $2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( 2\csc x \right)$, then the value of $x$ is(a) $\dfrac{3\pi }{4}$ (b) $\dfrac{\pi }{4}$ (c) $\dfrac{\pi }{3}$ (d) none of these

Hint: In inverse trigonometric functions, we have a formula using which we can add two ${{\tan }^{-1}}$ functions. The formula is, ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$.

Before proceeding with the question, we must know the formulas that are required to solve this question. In inverse trigonometric functions, we have a formula,
${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right).............\left( 1 \right)$
Also, in trigonometric functions, we have a formula,
$\csc x=\dfrac{1}{\sin x}.............\left( 2 \right)$
In this question, we have to solve the equation $2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( 2\csc x \right)$.
$\Rightarrow {{\tan }^{-1}}\left( \cos x \right)+{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( 2\csc x \right)...........\left( 3 \right)$
Using equation $\left( 1 \right)$ by substituting $a=\cos x$ and $b=\cos x$ from equation $\left( 3 \right)$, we get,
${{\tan }^{-1}}\left( \dfrac{\cos x+\cos x}{1-\cos x\cos x} \right)={{\tan }^{-1}}\left( 2\csc x \right)$
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{2\cos x}{1-{{\cos }^{2}}x} \right)={{\tan }^{-1}}\left( 2\csc x \right)............\left( 4 \right)$
Also, in trigonometric functions, we have an identity,
${{\sin }^{2}}x+{{\cos }^{2}}x=1..............\left( 5 \right)$
From equation $\left( 5 \right)$, we can also write,
$1-{{\cos }^{2}}x={{\sin }^{2}}x.................\left( 6 \right)$
Substituting $1-{{\cos }^{2}}x={{\sin }^{2}}x$ from equation $\left( 6 \right)$ in equation $\left( 4 \right)$, we get,
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{2\cos x}{{{\sin }^{2}}x} \right)={{\tan }^{-1}}\left( 2\csc x \right).............\left( 7 \right)$
Since in equation $\left( 7 \right)$, we have ${{\tan }^{-1}}$ on both sides of the equality.
Hence, we can now equate the arguments of ${{\tan }^{-1}}$ function in equation $\left( 7 \right)$.
$\Rightarrow \dfrac{2\cos x}{{{\sin }^{2}}x}=2\csc x..................\left( 8 \right)$
From equation $\left( 2 \right)$, we have $\csc x=\dfrac{1}{\sin x}$. Substituting $\csc x=\dfrac{1}{\sin x}$ from equation $\left( 2 \right)$ in equation $\left( 8 \right)$, we get,
$\dfrac{2\cos x}{{{\sin }^{2}}x}=\dfrac{2}{\sin x}............\left( 9 \right)$
Cancelling $2$ and $\sin x$ on both the sides of equality in equation $\left( 9 \right)$, we get,
\begin{align} & \dfrac{\cos x}{\sin x}=1 \\ & \Rightarrow \cos x=\sin x \\ \end{align}
Since, $\cos x$ is equal to $\sin x$ for $x=\dfrac{\pi }{4}$, therefore the answer is $x=\dfrac{\pi }{4}$.
Hence the answer is option (b).

Note: There can be more than one answer for this question. Since we had to solve the equation $\cos x=\sin x$ in the final step, we found it’s solution as $x=\dfrac{\pi }{4}$. But we must know that $\cos x=\sin x$ also for $x=\dfrac{5\pi }{4},\dfrac{9\pi }{4},\dfrac{13\pi }{4}......$. So, we must check for all the other options since there can be more than one option correct in a question. In the options of this question, there is only one option which is satisfying the equation $\cos x=\sin x$. That is why we have marked only a single option as a correct option in this question.