Question

If an inverse trigonometric equation is given as ${{\cos }^{-1}}x-{{\sin }^{-1}}x=0$ , then $x$ is equal to.
(a) $\pm \dfrac{1}{\sqrt{2}}$
(b) $1$
(c) $\pm \dfrac{1}{\sqrt{3}}$
(d) $\dfrac{1}{\sqrt{2}}$

Answer 1

Hint: For solving this question first we will prove an important formula of inverse trigonometric functions i.e., ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$. After that, we will find the angle ${{\sin }^{-1}}x$ and then solve further to find the exact value of $x$.

Complete step-by-step solution -
Given:
It is given that, ${{\cos }^{-1}}x-{{\sin }^{-1}}x=0$ and we have to find the value of $x$.
Now, let $\sin \theta =x$. Then,
$\begin{align}
  & \sin \theta =x \\
 & \Rightarrow \theta ={{\sin }^{-1}}x..................\left( 1 \right) \\
\end{align}$
Now, as we know that range of function $\theta ={{\sin }^{-1}}x$ will be from $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ . Then,
$-\dfrac{\pi }{2}\le \theta \le \dfrac{\pi }{2}...................\left( 2 \right)$
Now, we know that $\sin \theta =\cos \left( \dfrac{\pi }{2}-\theta \right)$ always. Then,
$\begin{align}
  & \sin \theta =x \\
 & \Rightarrow \cos \left( \dfrac{\pi }{2}-\theta \right)=x \\
\end{align}$
Now, let $\dfrac{\pi }{2}-\theta =\alpha $ so, we can write $\cos \left( \dfrac{\pi }{2}-\theta \right)=\cos \alpha $ in the above equation. Then,
$\begin{align}
  & \cos \left( \dfrac{\pi }{2}-\theta \right)=x \\
 & \Rightarrow \cos \alpha =x \\
 & \Rightarrow \alpha ={{\cos }^{-1}}x.................\left( 3 \right) \\
\end{align}$
Now, as we know that range of function $\alpha ={{\cos }^{-1}}x$ will be from $\left[ 0,\pi \right]$ . Then,
$0\le \alpha \le \pi $
Now, as per our assumption $\dfrac{\pi }{2}-\theta =\alpha $ so, we can write $\alpha =\dfrac{\pi }{2}-\theta $ in the above inequality. Then,
$\begin{align}
  & 0\le \alpha \le \pi \\
 & \Rightarrow 0\le \dfrac{\pi }{2}-\theta \le \pi \\
\end{align}$
Now, subtract $\dfrac{\pi }{2}$ in the above inequality. Then,
$\begin{align}
  & 0\le \dfrac{\pi }{2}-\theta \le \pi \\
 & \Rightarrow -\dfrac{\pi }{2}\le \dfrac{\pi }{2}-\theta -\dfrac{\pi }{2}\le \pi -\dfrac{\pi }{2} \\
 & \Rightarrow -\dfrac{\pi }{2}\le -\theta \le \dfrac{\pi }{2} \\
\end{align}$
Now, multiply by -1 in the above inequality. Then,
$\begin{align}
  & -\dfrac{\pi }{2}\le -\theta \le \dfrac{\pi }{2} \\
 & \Rightarrow \dfrac{\pi }{2}\ge \theta \ge -\dfrac{\pi }{2} \\
\end{align}$
Now, we can write the above inequality as $-\dfrac{\pi }{2}\le -\theta \le \dfrac{\pi }{2}$ and as this result matches with the equation (2) so, we conclude that we can write $\alpha ={{\cos }^{-1}}x$ for $\dfrac{\pi }{2}-\theta =\alpha $ always. Then,
$\begin{align}
  & \alpha ={{\cos }^{-1}}x \\
 & \Rightarrow \dfrac{\pi }{2}-\theta ={{\cos }^{-1}}x \\
\end{align}$
Now, put the value of $\theta ={{\sin }^{-1}}x$ from equation (1) in the above equation. Then,
$\begin{align}
  & \dfrac{\pi }{2}-\theta ={{\cos }^{-1}}x \\
 & \Rightarrow \dfrac{\pi }{2}-{{\sin }^{-1}}x={{\cos }^{-1}}x \\
 & \Rightarrow {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} \\
\end{align}$
Now, from the above result, we conclude that ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$ is always true. Then,
$\begin{align}
  & {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} \\
 & \Rightarrow {{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x \\
\end{align}$
Now, subtract ${{\sin }^{-1}}x$ from the above equation. Then,
$\begin{align}
  & {{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x \\
 & \Rightarrow {{\cos }^{-1}}x-{{\sin }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x-{{\sin }^{-1}}x \\
 & \Rightarrow {{\cos }^{-1}}x-{{\sin }^{-1}}x=\dfrac{\pi }{2}-2{{\sin }^{-1}}x \\
\end{align}$
Now, it is given that ${{\cos }^{-1}}x-{{\sin }^{-1}}x=0$ . Then,
$\begin{align}
  & {{\cos }^{-1}}x-{{\sin }^{-1}}x=\dfrac{\pi }{2}-2{{\sin }^{-1}}x \\
 & \Rightarrow 0=\dfrac{\pi }{2}-2{{\sin }^{-1}}x \\
 & \Rightarrow 2{{\sin }^{-1}}x=\dfrac{\pi }{2} \\
 & \Rightarrow {{\sin }^{-1}}x=\dfrac{\pi }{4} \\
\end{align}$
Now, apply sine operator in the above equation and then, write $\sin \left( {{\sin }^{-1}}x \right)=x$ and $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ . Then,
$\begin{align}
  & {{\sin }^{-1}}x=\dfrac{\pi }{4} \\
 & \Rightarrow \sin \left( {{\sin }^{-1}}x \right)=\sin \dfrac{\pi }{4} \\
 & \Rightarrow x=\dfrac{1}{\sqrt{2}} \\
\end{align}$
Now, from the above result, we conclude that if ${{\cos }^{-1}}x-{{\sin }^{-1}}x=0$ , then the value of $x=\dfrac{1}{\sqrt{2}}$ .
Hence, option (d) will be the correct option.

Note: Here, the student should first understand and then proceed in the right direction to get the correct answer quickly. Moreover, for objective problems, we should remember the result ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$ and apply it directly without any mistake to get the correct answer quickly without doing lengthy calculations.