Inverse Functions - Practice Questions & MCQ

Function f (f^-1 ( x )), where f(x) is a trigonometric function

1. $\sin \left(\sin ^{-1}(x)\right)=x \quad$ for all $x \in[-1,1]$
2. $\cos \left(\cos ^{-1}(x)\right)=x \quad$ for all $x \in[-1,1]$
3. $\tan \left(\tan ^{-1}(x)\right)=x \quad$ for all $x \in \mathbb{R}$
4. $\cot \left(\cot ^{-1}(x)\right)=x \quad$ for all $x \in \mathbb{R}$
5. $\sec \left(\sec ^{-1}(x)\right)=x \quad$ for all $x \in \mathbb{R}-(-1,1)$
6. $\csc \left(\csc ^{-1}(\mathrm{x})\right)=\mathrm{x} \quad$ for all $\mathrm{x} \in \mathbb{R}-(-1,1)$

The sine function is defined for all real values of $x$ but $\sin ^{-1}(x)$ is defined only for $x \in[-1,1]$.
Therefore, the domain of $\sin \left(\sin ^{-1}(x)\right)$ is $[-1,1]$.
Let $\theta \in[-\pi / 2, \pi / 2]$ and $x \in[-1,1]$, such that $\sin \theta=x$
Then, $\theta=\sin ^{-1} x$
Putting this value of $\theta$ in (i), we get, $x=\sin \left(\sin ^{-1} x\right)$
So, $\sin \left(\sin ^{-1} x\right)=x$ for all $x \in[-1,1]$
Similarly, $\cos \left(\cos ^{-1} x\right)=x$
Example

$
\begin{array}{ll}
\sin \left(\sin ^{-1}(1)\right)=\sin (\pi / 2)=1 & {\left[\because \sin ^{-1}(1)=\pi / 2\right]} \\
\sin \left(\sin ^{-1}(1 / 2)\right)=\sin (\pi / 6)=1 / 2 & {\left[\because \sin ^{-1}(1 / 2)=\pi / 6\right]} \\
\cos \left(\cos ^{-1}(1 / 2)\right)=\cos (\pi / 3)=1 / 2 & {\left[\because \cos ^{-1}(1 / 2)=\pi / 3\right]}
\end{array}
$
In the same way can prove other results
The graph of $f(x)=\sin \left(\sin ^{-1} x\right)$ or $f(x)=\cos \left(\cos ^{-1} x\right)$ is part of the line $y=x$ for $x \in[-1,1]$.

The graph of $f(x)=\tan \left(\tan ^{-1} x\right)$ or $f(x)=\cot \left(\cot ^{-1} x\right)$ for every $x \in R$ is the line $y=x$.

The graph of $f(x)=\sec \left(\sec ^{-1} x\right)$ or $f(x)=\operatorname{cosec}\left(\operatorname{cosec}^{-1} x\right)$ for every $x \in R-(-1,1)$ is a line $y=x$ but the part where x lies in $(-1,1)$ is deleted from it.

Principal Value of function f^-1 (f (x))

1. $\sin ^{-1}(\sin (\theta))=\theta \quad$ for all $\theta \in[-\pi / 2, \pi / 2]$
2. $\cos ^{-1}(\cos (\theta))=\theta \quad$ for all $\theta \in[0, \pi]$
3. $\tan ^{-1}(\tan (\theta))=\theta \quad$ for all $\theta \in(-\pi / 2, \pi / 2)$
4. $\cot ^{-1}(\cot (\theta))=\theta \quad$ for all $\theta \in(0, \pi)$
5. $\sec ^{-1}(\sec (\theta))=\theta \quad$ for all $\theta \in[0, \pi]-\{\pi / 2\}$
6. $\csc ^{-1}(\csc (\theta))=\theta \quad$ for all $\theta \in[-\pi / 2, \pi / 2]-\{0\}$

Let $\sin \theta=\mathrm{x}$,
Then, $\theta=\sin ^{-1} \mathrm{x}$
(As $\theta \in[-\pi / 2, \pi / 2]$ )
(Note that we cannot write $\theta=\sin ^{-1} x$ if $\theta$ does not lie in $[-\pi / 2, \pi / 2]$, as the value of $\sin ^{-1} x$ can only $b$ equal to an angle that lies in $[-\pi / 2, \pi / 2]$ )

So, $\theta=\sin ^{-1}(\sin \theta)$
[Replacing x with $\sin \theta$ ]
Hence, $\sin ^{-1}(\sin \theta)=\theta$
Hence, $\sin ^{-1}(\sin \theta)$ is true only for $\theta \in[-\pi / 2, \pi / 2]$
In the same way, we can prove other results.