Answer
Verified
408.6k+ views
Hint: Use the fact that if $y={{\cos }^{-1}}x$, then $x=\cos y$. Assume $y={{\cos }^{-1}}\left( \dfrac{1}{x} \right)$. Use the previously mentioned fact and write x in terms of y. Take ${{\sec }^{-1}}$ on both side and use the fact that if $y\in \left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$, then ${{\sec }^{-1}}\left( \sec y \right)=y$ and hence prove the result.
Complete step-by-step answer:
Before dwelling into the proof of the above question, we must understand how ${{\cos }^{-1}}x$ is defined even when $\cos x$ is not one-one.
We know that cosx is a periodic function.
Let us draw the graph of cosx
As is evident from the graph cosx is a repeated chunk of the graph of cosx within the interval $\left[ A,B \right]$ , and it attains all its possible values in the interval $\left[ A,C \right]$. Here $A=0,B=2\pi $ and $C=\pi $
Hence if we consider cosx in the interval [A, C], we will lose no value attained by cosx, and at the same time, cosx will be one-one and onto.
Hence ${{\cos }^{-1}}x$ is defined over the domain $\left[ -1,1 \right]$, with codomain $\left[ 0,\pi \right]$ as in the domain $\left[ 0,\pi \right]$, cosx is one-one and ${{R}_{\cos x}}=\left[ -1,1 \right]$.
Now since $\arccos x$ is the inverse of cosx it satisfies the fact that if $y=\arccos x$, then $\cos y=x$.
So let $y={{\cos }^{-1}}\left( \dfrac{1}{x} \right)$, $y\in \left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$ as $\dfrac{1}{x}\ne 0$
Hence we have $\cos y=\dfrac{1}{x}$
Hence we have $x=\dfrac{1}{\cos y}=\sec y$
Taking ${{\sec }^{-1}}$ on both sides, we get
${{\sec }^{-1}}x={{\sec }^{-1}}\left( \sec y \right)$
Now since $y\in \left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$ and we know that if $y\in \left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$, then ${{\sec }^{-1}}\left( \sec y \right)=y$ [Valid only in principal branch]
Hence we have
$\begin{align}
& {{\sec }^{-1}}x=y \\
& \Rightarrow y={{\sec }^{-1}}y \\
\end{align}$
Reverting to the original variable, we get
${{\cos }^{-1}}\left( \dfrac{1}{x} \right)={{\sec }^{-1}}\left( x \right)$
Since $\dfrac{1}{x}$ is in the domain of ${{\cos }^{-1}}x$, 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 ${{\cos }^{-1}}\left( \dfrac{1}{x} \right)={{\sec }^{-1}}x,x\in \left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)$
Note: [1] The above-specified codomain for ${{\cos }^{-1}}x$ is called principal branch for${{\cos }^{-1}}x$ . We can select any branch as long as $\cos x$ is one-one and onto and Range $=\left[ -1,1 \right]$. Like instead of $\left[ 0,\pi \right]$, we can select the interval $\left[ \pi ,2\pi \right]$. The proof will remain the same as above.
Complete step-by-step answer:
Before dwelling into the proof of the above question, we must understand how ${{\cos }^{-1}}x$ is defined even when $\cos x$ is not one-one.
We know that cosx is a periodic function.
Let us draw the graph of cosx
As is evident from the graph cosx is a repeated chunk of the graph of cosx within the interval $\left[ A,B \right]$ , and it attains all its possible values in the interval $\left[ A,C \right]$. Here $A=0,B=2\pi $ and $C=\pi $
Hence if we consider cosx in the interval [A, C], we will lose no value attained by cosx, and at the same time, cosx will be one-one and onto.
Hence ${{\cos }^{-1}}x$ is defined over the domain $\left[ -1,1 \right]$, with codomain $\left[ 0,\pi \right]$ as in the domain $\left[ 0,\pi \right]$, cosx is one-one and ${{R}_{\cos x}}=\left[ -1,1 \right]$.
Now since $\arccos x$ is the inverse of cosx it satisfies the fact that if $y=\arccos x$, then $\cos y=x$.
So let $y={{\cos }^{-1}}\left( \dfrac{1}{x} \right)$, $y\in \left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$ as $\dfrac{1}{x}\ne 0$
Hence we have $\cos y=\dfrac{1}{x}$
Hence we have $x=\dfrac{1}{\cos y}=\sec y$
Taking ${{\sec }^{-1}}$ on both sides, we get
${{\sec }^{-1}}x={{\sec }^{-1}}\left( \sec y \right)$
Now since $y\in \left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$ and we know that if $y\in \left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$, then ${{\sec }^{-1}}\left( \sec y \right)=y$ [Valid only in principal branch]
Hence we have
$\begin{align}
& {{\sec }^{-1}}x=y \\
& \Rightarrow y={{\sec }^{-1}}y \\
\end{align}$
Reverting to the original variable, we get
${{\cos }^{-1}}\left( \dfrac{1}{x} \right)={{\sec }^{-1}}\left( x \right)$
Since $\dfrac{1}{x}$ is in the domain of ${{\cos }^{-1}}x$, 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 ${{\cos }^{-1}}\left( \dfrac{1}{x} \right)={{\sec }^{-1}}x,x\in \left( -\infty ,-1 \right]\bigcup \left[ 1,\infty \right)$
Note: [1] The above-specified codomain for ${{\cos }^{-1}}x$ is called principal branch for${{\cos }^{-1}}x$ . We can select any branch as long as $\cos x$ is one-one and onto and Range $=\left[ -1,1 \right]$. Like instead of $\left[ 0,\pi \right]$, we can select the interval $\left[ \pi ,2\pi \right]$. The proof will remain the same as above.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE