Question

Prove that ${{\sin }^{-1}}\left( \dfrac{1}{x} \right)={{\csc }^{-1}}x,x\in \left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)$

Answer 1

Hint: Use the fact that if $y={{\sin }^{-1}}x$, then $x=\sin y$. Assume $y={{\sin }^{-1}}\left( \dfrac{1}{x} \right)$. Use the previously mentioned fact and write x in terms of y. ]. Take ${{\sec }^{-1}}$ on both sides and use the fact that if y\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\}, then ${{\csc }^{-1}}\left( \csc y \right)=y$ and hence prove the result.

Complete step-by-step answer:
Before dwelling into the proof of the above question, we must understand how ${{\sin }^{-1}}x$ is defined even when $\sin x$ is not one-one.
We know that sinx is a periodic function.
Let us draw the graph of sinx

As is evident from the graph sinx is a repeated chunk of the graph of sinx within the interval $\left[ A,B \right]$ , and it attains all its possible values in the interval $\left[ A,C \right]$. Here $A=\dfrac{-\pi }{2},B=\dfrac{3\pi }{2}$ and $C=\dfrac{\pi }{2}$
Hence if we consider sinx in the interval [A, C], we will lose no value attained by sinx, and at the same time, sinx will be one-one and onto.
Hence ${{\sin }^{-1}}x$ is defined over the domain $\left[ -1,1 \right]$, with codomain $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ as in the domain $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$, sinx is one-one and ${{R}_{\sin x}}=\left[ -1,1 \right]$.
Now since ${{\sin }^{-1}}x$ is the inverse of sinx it satisfies the fact that if $y={{\sin }^{-1}}x$, then $\sin y=x$.
So let $y={{\sin }^{-1}}\left( \dfrac{1}{x} \right)$, y\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\} as $\dfrac{1}{x}\ne 0$
Hence we have $\sin y=\dfrac{1}{x}$
Hence we have $x=\dfrac{1}{\sin y}=\csc y$
Taking ${{\csc }^{-1}}$ on both sides, we get
${{\csc }^{-1}}x={{\csc }^{-1}}\left( \csc y \right)$
Now since y\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\} and we know that if y\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\}, then ${{\csc }^{-1}}\left( \csc y \right)=y$ [Valid only in principal branch]
Hence we have
$\begin{aligned} & {{\csc }^{-1}}x=y \\\ & \Rightarrow y={{\csc }^{-1}}y \\\ \end{aligned}$
Reverting to the original variable, we get
${{\sin }^{-1}}\left( \dfrac{1}{x} \right)={{\csc }^{-1}}\left( x \right)$
Since $\dfrac{1}{x}$ is in the domain of ${{\sin }^{-1}}x$, we get
$\dfrac{1}{x}\in \left[ -1,1 \right]\Rightarrow x\in \left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)$
Hence we have ${{\sin }^{-1}}\left( \dfrac{1}{x} \right)={{\csc }^{-1}}x,x\in \left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)$

Note: [1] The above-specified codomain for ${{\sin }^{-1}}x$ is called principal branch for ${{\sin }^{-1}}x$. We can select any branch as long as $\sin x$ is one-one and onto and Range $=\left[ -1,1 \right]$. Like instead of $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, we can select the interval $\left[ \dfrac{\pi }{2},\dfrac{3\pi }{2} \right]$. The proof will remain the same as above.