Answer
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Hint: The Inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, etc. The inverse trigonometric functions are used to find the angle measure of a right-angled triangle when the measure of two sides of the triangle are known. The conventional symbol used to represent them is ‘arcsin’, ‘arccosine’, ‘arctan’, etc.
Complete step by step answer:
We will now see the domain and range of all the six inverse trigonometric functions in the following order:
(1) ${{\sin }^{-1}}\left( x \right)$
The domain of ${{\sin }^{-1}}\left( x \right)$ is equal to the range of $\sin \left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ {{\sin }^{-1}}(x) \right]=\left[ -1,1 \right]$
And, the range of ${{\sin }^{-1}}\left( x \right)$ is equal to the domain of $\sin \left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ {{\sin }^{-1}}\left( x \right) \right]=\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$
(2) ${{\cos }^{-1}}\left( x \right)$
The domain of ${{\cos }^{-1}}\left( x \right)$ is equal to the range of $\cos \left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ {{\cos }^{-1}}(x) \right]=\left[ -1,1 \right]$
And, the range of ${{\cos }^{-1}}\left( x \right)$ is equal to the domain of $\cos \left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ {{\cos }^{-1}}\left( x \right) \right]=\left[ 0,\pi \right]$
(3) ${{\tan }^{-1}}\left( x \right)$
The domain of ${{\tan }^{-1}}\left( x \right)$ is equal to the range of $\tan \left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ {{\tan }^{-1}}(x) \right]=\left( -\infty ,\infty \right)$
And, the range of ${{\tan }^{-1}}\left( x \right)$ is equal to the domain of $\tan \left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ {{\tan }^{-1}}\left( x \right) \right]=\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$
(4) ${{\cot }^{-1}}\left( x \right)$
The domain of ${{\cot }^{-1}}\left( x \right)$ is equal to the range of $\cot \left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ {{\cot }^{-1}}(x) \right]=\left( -\infty ,\infty \right)$
And, the range of ${{\cot }^{-1}}\left( x \right)$ is equal to the domain of $\cot \left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ {{\cot }^{-1}}\left( x \right) \right]=\left( 0,\pi \right)$
(5) $\cos e{{c}^{-1}}\left( x \right)$
The domain of $\cos e{{c}^{-1}}\left( x \right)$ is equal to the range of $\cos ec\left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ \cos e{{c}^{-1}}(x) \right]=(-\infty ,1]\cup [1,\infty )$
And, the range of $\cos e{{c}^{-1}}\left( x \right)$ is equal to the domain of $\cos ec\left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ \cos e{{c}^{-1}}\left( x \right) \right]=\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]-\left\{ 0 \right\}$
(6) ${{\sec }^{-1}}\left( x \right)$
The domain of ${{\sec }^{-1}}\left( x \right)$ is equal to the range of $\sec \left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ {{\sec }^{-1}}(x) \right]=(-\infty ,1]\cup [1,\infty )$
And, the range of ${{\sec }^{-1}}\left( x \right)$ is equal to the domain of $\sec \left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ {{\sec }^{-1}}\left( x \right) \right]=\left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$
Note: The inverse functions are basically the mirror image of the fundamental functions. That is, they are identical in shape about the line, $y=x$ . This property is used in problems to plot the graph of these inverse trigonometric functions.
Complete step by step answer:
We will now see the domain and range of all the six inverse trigonometric functions in the following order:
(1) ${{\sin }^{-1}}\left( x \right)$
The domain of ${{\sin }^{-1}}\left( x \right)$ is equal to the range of $\sin \left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ {{\sin }^{-1}}(x) \right]=\left[ -1,1 \right]$
And, the range of ${{\sin }^{-1}}\left( x \right)$ is equal to the domain of $\sin \left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ {{\sin }^{-1}}\left( x \right) \right]=\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$
(2) ${{\cos }^{-1}}\left( x \right)$
The domain of ${{\cos }^{-1}}\left( x \right)$ is equal to the range of $\cos \left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ {{\cos }^{-1}}(x) \right]=\left[ -1,1 \right]$
And, the range of ${{\cos }^{-1}}\left( x \right)$ is equal to the domain of $\cos \left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ {{\cos }^{-1}}\left( x \right) \right]=\left[ 0,\pi \right]$
(3) ${{\tan }^{-1}}\left( x \right)$
The domain of ${{\tan }^{-1}}\left( x \right)$ is equal to the range of $\tan \left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ {{\tan }^{-1}}(x) \right]=\left( -\infty ,\infty \right)$
And, the range of ${{\tan }^{-1}}\left( x \right)$ is equal to the domain of $\tan \left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ {{\tan }^{-1}}\left( x \right) \right]=\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$
(4) ${{\cot }^{-1}}\left( x \right)$
The domain of ${{\cot }^{-1}}\left( x \right)$ is equal to the range of $\cot \left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ {{\cot }^{-1}}(x) \right]=\left( -\infty ,\infty \right)$
And, the range of ${{\cot }^{-1}}\left( x \right)$ is equal to the domain of $\cot \left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ {{\cot }^{-1}}\left( x \right) \right]=\left( 0,\pi \right)$
(5) $\cos e{{c}^{-1}}\left( x \right)$
The domain of $\cos e{{c}^{-1}}\left( x \right)$ is equal to the range of $\cos ec\left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ \cos e{{c}^{-1}}(x) \right]=(-\infty ,1]\cup [1,\infty )$
And, the range of $\cos e{{c}^{-1}}\left( x \right)$ is equal to the domain of $\cos ec\left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ \cos e{{c}^{-1}}\left( x \right) \right]=\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]-\left\{ 0 \right\}$
(6) ${{\sec }^{-1}}\left( x \right)$
The domain of ${{\sec }^{-1}}\left( x \right)$ is equal to the range of $\sec \left( x \right)$. So, it could be written as:
$\Rightarrow D\left[ {{\sec }^{-1}}(x) \right]=(-\infty ,1]\cup [1,\infty )$
And, the range of ${{\sec }^{-1}}\left( x \right)$ is equal to the domain of $\sec \left( x \right)$. So, it could be written as:
$\Rightarrow R\left[ {{\sec }^{-1}}\left( x \right) \right]=\left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}$
Note: The inverse functions are basically the mirror image of the fundamental functions. That is, they are identical in shape about the line, $y=x$ . This property is used in problems to plot the graph of these inverse trigonometric functions.
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