Question

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

Answer 1

Hint: Use the fact that if $y={{\sec }^{-1}}x$, then $x=\sec y$. Assume $y={{\sec }^{-1}}$. Write $\sec \left( {{\sec }^{-1}}x \right)$ in terms of y and hence prove the above result.

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

As is evident from the graph secx is a repeated chunk of the graph of secx within the interval \left[ A,B \right]-\left\\{ \dfrac{\pi }{2},\dfrac{3\pi }{2} \right\\} , and it attains all its possible values in the interval \left[ A,C \right]-\left\\{ \dfrac{\pi }{2} \right\\}.
Here $A=0,B=2\pi$ and $C=\pi$
Hence if we consider secx in the interval \left[ A,C \right]-\left\\{ \dfrac{\pi }{2} \right\\}, we will lose no value attained by secx, and at the same time, secx will be one-one and onto.
Hence ${{\sec }^{-1}}x$ is defined over the Domain $\left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)$, with codomain \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\} as in the Domain \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\}, secx is one-one and ${{R}_{\sec x}}=\left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)$.
Now since ${{\sec }^{-1}}x$ is the inverse of secx it satisfies the fact that if $y={{\sec }^{-1}}x$, then $\sec y=x$.
So let $y={{\sec }^{-1}}x$
Hence we have secy = x.
Now $\sec \left( {{\sec }^{-1}}x \right)=\sec \left( y \right)$
Hence we have $\sec \left( {{\sec }^{-1}}x \right)=x$.
Also as x is in the Domain of ${{\sec }^{-1}}x$, we have $x\in \left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)$.
Hence $\sec \left( {{\sec }^{-1}}x \right)=x,x\in \left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)$

Note: [1] The above-specified codomain for ${{\sec }^{-1}}x$ is called principal branch for ${{\sec }^{-1}}x$. We can select any branch as long as $\sec x$ is one-one and onto and Range $=\left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)$. Like instead of \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\}, we can select the interval \left[ \pi ,2\pi \right]-\left\\{ \dfrac{3\pi }{2} \right\\}. The proof will remain the same as above.