Graph of Inverse Trigonometric Function - Practice Questions & MCQ

Graph of Principal Value of function $\mathrm{f}^{-1}(\mathrm{f}(\mathrm{x}))(\sin$ and $\cos )$
Graph of $\sin ^{-1}(\sin (x))$

$
\begin{aligned}
& y=\sin ^{-1}(\sin (x)) \\
& \sin y=\sin x
\end{aligned}
$
$
\Rightarrow \quad y=n \pi+(-1)^n x, n \in I \text { (integer) }
$
Since, $y \in[-\pi / 2, \pi / 2]$, we have different expressions for $\sin ^{-1}(\sin (x))$ for different values of $x$.

From above, we can plot the graph of $y=\sin -1(\sin (x))$

2. Graph of $\cos ^{-1}(\cos (\mathrm{x}))$

$
\begin{array}{ll} 
& y=\cos ^{-1}(\cos (x)) \\
\therefore & \cos y=\cos x \\
\Rightarrow & y=2 n \pi \pm x, n \in I \text { (integer) }
\end{array}
$
To draw the graph of $y=\cos ^{-1}(\cos (x))$, we draw all the lines of $y=2 n \pi \pm x, n \in I$ for $y \in[0, \pi]$.

Graph of Principal Value of function $\mathrm{f}^{-1}(\mathrm{f}(\mathrm{x})$ ) (tan and cot)
1. Graph of $\tan ^{-1}(\tan (\mathrm{x}))$

The domain of the function is R and range is $(-\pi / 2, \pi / 2)$

$
\begin{array}{ll} 
& \mathrm{y}=\tan ^{-1}(\tan (\mathrm{x})) \\
\therefore & \tan \mathrm{y}=\tan \mathrm{x} \\
\Rightarrow & \mathrm{y}=\mathrm{n} \pi+\mathrm{x}, \mathrm{n} \in \mathrm{I} \text { (integer) }
\end{array}
$
To draw the graph of $y=\tan ^{-1}(\tan (x))$, we draw all the line of $y=n \pi+x, n \in I$ for $y \in(-\pi / 2, \pi / 2)$

2. Graph of $\cot ^{-1}(\cot (x))$

The domain of the function is $R$ and the range is $(0, \pi)$

$
\begin{array}{ll} 
& y=\cot ^{-1}(\cot (x)) \\
\therefore & \cot y=\cot x \\
\Rightarrow & y=n \pi+x, n \in I \text { (integer) }
\end{array}
$
To draw the graph of $y=\cot ^{-1}(\cot (x))$, we draw all the lines of $y=n \pi+x, n \in I$ for $y \in(0, \pi)$.

Graph of Principal Value of function $f^{-1}(f(x))(\operatorname{cosec}$ and $s e c)$
1. Graph of $\operatorname{cosec}^{-1}(\operatorname{cosec}(x))$

Domain of the function is $R-\{n \pi, n \in l\}$ and range is $[-\pi / 2, \pi / 2]-\{0\}$

$
\begin{array}{ll} 
& y=\operatorname{cosec}^{-1}(\operatorname{cosec}(x)) \\
\therefore \quad & \operatorname{cosec} y=\operatorname{cosec} x \\
\text { or } & \sin x=\sin y
\end{array}
$
Hence, the graph of $\operatorname{cosec}^{-1}(\operatorname{cosec}(x))$ is the same as that of $y=\sin ^{-1}(\sin (x))$, but excluding points $\mathrm{x}=\mathrm{n} \pi, \mathrm{n} \in \mathrm{I}$.

2. Graph of $\sec ^{-1}(\sec (x))$

Domain of the function is $R-\{(2 n+1) \pi / 2, n \in l\}$ and range is $[0, \pi]-\{\pi / 2\}$

$
\begin{array}{ll} 
& y=\sec ^{-1}(\sec (x)) \\
\therefore \quad & \sec y=\sec x \\
\text { or } \quad & \cos y=\cos x
\end{array}
$
Hence, graph of $\sec ^{-1}(\sec (x))$ is the same as that of $y=\cos ^{-1}(\cos (x))$, but excluding points $x=$ $(2 n+1) \pi / 2, n \in I$.

Conversion of one ITF to another
We can write the different inverse trigonometric functions in terms of other inverse trigonometric functions.
1. Converting $\sin ^{-1} x$

If $0<x<1$
Let $\sin ^{-1} \mathrm{x}=\Theta$. Then $\sin \Theta=\mathrm{x}$ or $\sin \Theta=\mathrm{x} / 1$
So, we have the right-angled triangle

From the figure

$
\begin{aligned}
& \cos \theta=\frac{\sqrt{1-\mathrm{x}^2}}{1}=\sqrt{1-\mathrm{x}^2} \\
& \therefore \quad \theta=\sin ^{-1} \mathrm{x}=\cos ^{-1} \sqrt{1-\mathrm{x}^2}
\end{aligned}
$
Also, we have $\theta=\tan ^{-1} \frac{x}{\sqrt{1-x^2}}$

$
\begin{aligned}
& =\cot ^{-1} \frac{\sqrt{1-\mathrm{x}^2}}{\mathrm{x}} \\
& =\sec ^{-1} \frac{1}{\sqrt{1-\mathrm{x}^2}} \\
& =\csc ^{-1} \frac{1}{\mathrm{x}}
\end{aligned}
$

2. Converting cos^-1 x

If $0<x<1$
Let $\cos ^{-1} \mathrm{x}=\Theta$, implies $\cos \Theta=\mathrm{x}$
We have the following right-angled triangle

From the figure

$
\begin{aligned}
& \sin \theta=\frac{\sqrt{1-\mathrm{x}^2}}{1}=\sqrt{1-\mathrm{x}^2} \\
& \therefore \quad \theta=\cos ^{-1} \mathrm{x}=\sin ^{-1} \sqrt{1-\mathrm{x}^2}
\end{aligned}
$
Also, we have $\theta=\tan ^{-1} \frac{\sqrt{1-x^2}}{x}$

$
\begin{aligned}
& =\cot ^{-1} \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^2}} \\
& =\sec ^{-1} \frac{1}{\mathrm{x}} \\
& =\csc ^{-1} \frac{1}{\sqrt{1-\mathrm{x}^2}}
\end{aligned}
$

3. Converting $\tan ^{-1} x$

If $x>0$
Let $\tan ^{-1} \mathrm{x}=\Theta$, implies $\tan \Theta=\mathrm{x}$

We have the following right-angled triangle

From the figure

$
\begin{aligned}
\sin \theta & =\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}} \\
\theta & =\tan ^{-1} \mathrm{x}=\sin ^{-1} \frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}
\end{aligned}
$
Also, we have $\theta=\cos ^{-1} \frac{1}{\sqrt{1+x^2}}$

$
\begin{aligned}
& =\cot ^{-1} \frac{1}{x} \\
& =\sec ^{-1} \sqrt{1+x^2} \\
& =\csc ^{-1} \frac{\sqrt{1+x^2}}{x}
\end{aligned}
$

Relating $f^{-1}(x)$ with $f^{-1}(-x)$
We have the following important results
1. $\sin ^{-1}(-x)=-\sin ^{-1}(x)$ for all $x \in[-1,1]$

Proof:
Let $-x \in[-1,1]$, so $x \in[-1,1]$
Also assume $\sin ^{-1}(-x)=\Theta \quad$ (Hence $\left.\Theta \in[-\pi / 2, \pi / 2]\right)$
Taking sin of both sides

$
\begin{aligned}
& \sin \left(\sin ^{-1}(-x)\right)=\sin (\Theta) \\
& -x=\sin \Theta \\
& \Rightarrow x=-\sin \Theta \\
& \Rightarrow x=\sin (-\Theta)
\end{aligned}
$
Taking $\sin ^{-1}$ of both sides

$
\Rightarrow \sin ^{-1}(x)=\sin ^{-1}(\sin (-\Theta))
$

$[$ As $\Theta \in[-\pi / 2, \pi / 2]$, so $-\Theta \in[-\pi / 2, \pi / 2]]$

$
\Rightarrow-\sin ^{-1}(x)=\Theta
$

From (i) and (ii), we get

$
\Rightarrow \sin ^{-1}(-x)=-\sin ^{-1}(x)
$
Similarly, we can prove
2. $\tan ^{-1}(-x)=-\tan ^{-1}(x)$ for all $x \in R$ and
3. $\operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1}(x)$ for all $x \in R-(-1,1)$
4. $\cos ^{-1}(-x)=\pi-\cos ^{-1}(x)$ for all $x \in[-1,1]$

Proof:

Let $-x \in[-1,1]$
And assume $\cos ^{-1}(-x)=\Theta \quad$ (Hence $\Theta \in[0, \pi]$ )
Taking cos of both sides

$
\begin{aligned}
& \cos \left(\cos ^{-1}(-\mathrm{x})\right)=\cos \Theta \\
& \therefore-\mathrm{x}=\cos \Theta \\
& \Rightarrow \mathrm{x}=-\cos \Theta \\
& \Rightarrow \mathrm{x}=\cos (\pi-\Theta)
\end{aligned}
$
Taking $\cos ^{-1}$ of both sides

$
\begin{aligned}
& \Rightarrow \cos ^{-1}(x)=\cos ^{-1}(\cos (\pi-\Theta)) \\
& {[\because \Theta \in[0, \pi], \text { so } \pi-\Theta \in[0, \pi]]}
\end{aligned}
$

$
\Rightarrow \Theta=\pi-\cos ^{-1}(x)
$
From (i) and (ii), we get

$
\Rightarrow \cos ^{-1}(-x)=\pi-\cos ^{-1}(x)
$
Similarly, we can prove
5. $\sec ^{-1}(-x)=\pi-\sec ^{-1}(x)$ for all $x \in R-(-1,1)$
6. $\cot ^{-1}(-x)=\pi-\cot ^{-1}(x)$ for all $x \in R$

Relating $f^{-1}(x)$ with $f^{-1}(1 / x)$
1. $\sin ^{-1}\left(\frac{1}{\mathrm{x}}\right)=\csc ^{-1} \mathrm{x} \quad$ for all $\mathrm{x} \in(-\infty,-1] \cup[1, \infty)$
2. $\cos ^{-1}\left(\frac{1}{\mathrm{x}}\right)=\sec ^{-1} \mathrm{x} \quad$ for all $\mathrm{x} \in(-\infty,-1] \cup[1, \infty)$
3. $\tan ^{-1}\left(\frac{1}{\mathrm{x}}\right)=\left\{\begin{array}{cc}\cot ^{-1} x & \text { for } x>0 \\ -\pi+\cot ^{-1} x & \text { for } x<0\end{array}\right.$

Proof:
1.

$
\begin{aligned}
& \text { Let } \csc ^{-1} \mathrm{x}=\theta \\
& \text { where } \theta \in[-\pi / 2, \pi / 2]-\{0\} \\
& \text { and } \mathrm{x} \in(-\infty,-1] \cup[1, \infty) \\
& \therefore \quad \mathrm{x}=\csc \theta \\
& \Rightarrow \quad \frac{1}{\mathrm{x}}=\sin \theta \\
& \Rightarrow \quad \sin ^{-1} \frac{1}{\mathrm{x}}=\sin ^{-1}(\sin \theta) \\
& \Rightarrow \quad \theta=\sin ^{-1} \frac{1}{\mathrm{x}}(\text { as } \theta \in[-\pi / 2, \pi / 2]-\{0\})
\end{aligned}
$
From (i) and (ii) we get

$
\sin ^{-1}\left(\frac{1}{x}\right)=\csc ^{-1} x
$

2.

$
\begin{aligned}
& \text { Let } \sec ^{-1} \mathrm{x}=\theta \\
& \text { where } \theta \in[0, \pi]-\{\pi / 2\} \\
& \text { and } \mathrm{x} \in(-\infty,-1] \cup[1, \infty) \\
& \therefore \quad \mathrm{x}=\sec \theta \\
& \Rightarrow \quad \frac{1}{\mathrm{x}}=\cos \theta \\
& \Rightarrow \quad \cos ^{-1} \frac{1}{\mathrm{x}}=\cos ^{-1}(\cos \theta) \\
& \Rightarrow \quad \theta=\cos ^{-1} \frac{1}{\mathrm{x}}(\operatorname{as} \theta \in[0, \pi]-\{\pi / 2\})
\end{aligned}
$
From (i) and (ii) we get

$
\cos ^{-1}\left(\frac{1}{x}\right)=\sec ^{-1} x
$

3.

Let $\cot ^{-1} \mathrm{x}=\theta$, where $\theta \in(0, \pi)$ and $\mathrm{x} \in \mathrm{R}$

$
\begin{array}{ll}
\therefore & \mathrm{x}=\cot \theta \\
\Rightarrow & \frac{1}{\mathrm{x}}=\tan \theta \\
\Rightarrow & \tan ^{-1}\left(\frac{1}{\mathrm{x}}\right)=\tan ^{-1}(\tan \theta)
\end{array}
$

From the graph

$\begin{aligned} \tan ^{-1}\left(\frac{1}{\mathrm{x}}\right) & = \begin{cases}\theta, & 0<\theta<\pi / 2 \\ -\pi+\theta, & \pi / 2<\theta<\pi\end{cases} \\ & = \begin{cases}\cot ^{-1} x, & 0<\cot ^{-1} x<\pi / 2 \\ -\pi+\cot ^{-1} x, \pi / 2<\cot ^{-1} x<\pi\end{cases} \\ & = \begin{cases}\cot ^{-1} x, & x>0 \\ -\pi+\cot ^{-1} x, x<0\end{cases} \end{aligned}$