Question

Range of ${{\sin }^{-1}}x-{{\cos }^{-1}}x$ is:
A) $\left[ -\dfrac{3\pi }{2},\dfrac{\pi }{2} \right]$
B) $\left[ -\dfrac{5\pi }{3},\dfrac{\pi }{3} \right]$
C) $\left[ -\dfrac{3\pi }{2},\pi \right]$
D) $\left[ 0,\pi \right]$

Answer 1

First, start with the range of the inverse of sin is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ and of cos is $\left[ 0,\pi \right]$. After that add and subtract ${{\cos }^{-1}}x$ to get positive signs of ${{\cos }^{-1}}x$. As, we know that ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$, substitute $\dfrac{\pi }{2}$ in function. After that substitute the range of ${{\cos }^{-1}}x$ in the function and solve it further to get the range of the function.

Complete step-by-step answer:
Given:- $f\left( x \right)={{\sin }^{-1}}x-{{\cos }^{-1}}x$
As we know that the range of ${{\sin }^{-1}}x$ is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ and the range of ${{\cos }^{-1}}x$ is $\left[ 0,\pi \right]$.
Now, add and subtract ${{\cos }^{-1}}x$ in the function,
$\Rightarrow$ $f\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x-{{\cos }^{-1}}x-{{\cos }^{-1}}x$
As we know that ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$. Substitute $\dfrac{\pi }{2}$ in place of ${{\sin }^{-1}}x+{{\cos }^{-1}}x$,
$\Rightarrow$ $f\left( x \right)=\dfrac{\pi }{2}-2{{\cos }^{-1}}x$
Substitute the range of ${{\cos }^{-1}}x$ to find the range of the function,
$\Rightarrow$ $f\left( x \right)\in \left[ \dfrac{\pi }{2}-2\left( \pi \right),\dfrac{\pi }{2}-2\left( 0 \right) \right]$
Open the bracket and multiply the terms,
$f\left( x \right)\in \left[ \dfrac{\pi }{2}-2\pi ,\dfrac{\pi }{2}-0 \right]$
Subtract the terms,
$\Rightarrow$ $f\left( x \right)\in \left[ -\dfrac{3\pi }{2},\dfrac{\pi }{2} \right]$
Thus, the range of the function $f\left( x \right)$ is $\left[ -\dfrac{3\pi }{2},\dfrac{\pi }{2} \right]$.

Hence, option (C) is the correct answer.

Note: A function is a relation for which each value from the set of the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair.
The range of a function is the set of all possible outputs for the function.
Inverse trigonometric functions are also called “Arc Functions” since, for a given value of trigonometric functions, they produce the length of arc needed to obtain that particular value. The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent.
The domain and range of inverse trigonometry functions are: -

Function	Domain	Range
${{\sin }^{-1}}x$	[-1, 1]	$\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$
${{\cos }^{-1}}x$	[-1, 1]	$\left[ 0,\pi \right]$
${{\tan }^{-1}}x$	For all real numbers	$\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$
${{\cot }^{-1}}x$	For all real numbers	$\left( 0,\pi \right)$
${{\sec }^{-1}}x$	$\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)$	$\left[ 0,\dfrac{\pi }{2} \right)\cup \left( \dfrac{\pi }{2},\pi \right]$
${{\operatorname{cosec}}^{-1}}x$	$\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)$	$\left[ -\dfrac{\pi }{2},0 \right)\cup \left( 0,\dfrac{\pi }{2} \right]$.