Exercise 4.3 - Chapter 4 Inverse Trigonometric Functions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.3$
Question 1.
Find the domain of the following functions:
(i) $\tan ^{-1}\left(\sqrt{9-x^2}\right)$
(ii) $\frac{1}{2} \tan ^{-1}\left(1-x^2\right)-\frac{\pi}{4}$
Solution:
(i) $f(x)=\tan ^{-1}\left(\sqrt{9-x^2}\right)$
We know the domain of $\tan ^{-1} \mathrm{x}$ is $(-\infty, \infty)$ and range is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
So, the domain of $\mathrm{f}(\mathrm{x})=\tan ^{-1}\left(\sqrt{9-x^2}\right)$ is the set of values of $\mathrm{x}$ satisfying the inequality
$
\begin{aligned}
& -\infty \leq \sqrt{9-x^2} \leq \infty \\
& \Rightarrow 9-x^2 \geq 0 \\
& \Rightarrow x^2 \leq 9 \\
& \Rightarrow|x| \leq 3
\end{aligned}
$
Since $\tan \mathrm{x}$ is an odd function and symmetric about the origin, $\tan ^{-1} \mathrm{x}$ should be an increasing function in its domain.
$\therefore$ Domain is $(2 \mathrm{n}+1) \frac{\pi}{2}$
(ii) We know $\quad \frac{1}{2} \tan ^{-1} x=\tan ^{-1}\left(\frac{1-x}{1+x}\right)$
So
$
\frac{1}{2} \tan ^{-1}\left(1-x^2\right)=\tan ^{-1}\left[\frac{1-\left(1-x^2\right)}{1+\left(x-x^2\right)}\right]=\tan ^{-1} \frac{x^2}{2-x^2}
$
So
$
\begin{aligned}
\frac{1}{2} \tan ^{-1}\left(1-x^2\right)-\frac{\pi}{4} & =\tan ^{-1} \frac{x^2}{2-x^2}-\tan ^{-1} 1=\tan ^{-1}\left[\frac{\frac{x^2}{2-x^2}-1}{1+\left(\frac{x^2}{2-x^2}\right)(1)}\right] \\
& =\tan ^{-1}\left(x^2-1\right)=y \text { (say) }
\end{aligned}
$
The domain of $\mathrm{y}$ is $(-\infty, \infty)\{\mathrm{x} \mid \mathrm{x} \in-1\}$ and range is $[-1, \infty)\{\mathrm{y} \mid \mathrm{y} \geq-1\}$
The domain for $\tan ^{-1}\left(\mathrm{x}^2-1\right)$ is $(2 \mathrm{n}+1) \pi$. Since $\tan \mathrm{x}$ is an odd function.

Question 2.
Find the value of
(i) $\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right)$
(ii) $\tan ^{-1}\left(\tan \left(-\frac{\pi}{6}\right)\right)$
Solution:
(i)
$
\begin{aligned}
\tan \frac{5 \pi}{4} & =\tan \left(\pi+\frac{\pi}{4}\right)=\tan \frac{\pi}{4} \\
\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right) & =\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4}
\end{aligned}
$

(ii)
So
$
\begin{aligned}
\tan \left(-\frac{\pi}{6}\right) & =\tan \left(\pi-\frac{\pi}{6}\right)=\tan \frac{5 \pi}{6} \\
\tan ^{-1}\left(\tan \left(-\frac{\pi}{6}\right)\right) & =\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)=\frac{5 \pi}{6}
\end{aligned}
$

Question 3.
Find the value of
(i) $\tan \left(\tan ^{-1} \frac{7 \pi}{4}\right)$
(ii) $\tan \left(\tan ^{-1}(1947)\right)$
(iii) $\tan \left(\tan ^{-1}(-0.2021)\right)$
Solution:
We know that $\tan \left(\tan ^{-1} \mathrm{x}\right)=\mathrm{x}$
(i) $\tan \left(\tan ^{-1} \frac{7 \pi}{4}\right)=\frac{7 \pi}{4}$
(ii) $\tan \left(\tan ^{-1}(1947)\right)=1947$
(iii) $\tan \left(\tan ^{-1}(-0.2021)\right)=-0.2021$
Question 4.
Find the value of
(i) $\tan \left(\cos ^{-1}\left(\frac{1}{2}\right)-\sin ^{-1}\left(-\frac{1}{2}\right)\right)$
(ii) $\sin \left(\tan ^{-1}\left(\frac{1}{2}\right)-\cos ^{-1}\left(\frac{4}{5}\right)\right)$
(iii) $\cos \left(\sin ^{-1}\left(\frac{4}{5}\right)-\tan ^{-1}\left(\frac{3}{4}\right)\right)$
Solution:

(i)
$
\begin{aligned}
\cos ^{-1}\left(\frac{1}{2}\right) & =\frac{\pi}{3} \\
\sin ^{-1}\left(-\frac{1}{2}\right) & =-\frac{\pi}{6}
\end{aligned}
$
So $\quad \cos ^{-1}\left(\frac{1}{2}\right)-\sin ^{-1}\left(-\frac{1}{2}\right)=\frac{\pi}{3}-\left(\frac{-\pi}{6}\right)=\frac{\pi}{3}+\frac{\pi}{6}=\frac{2 \pi+\pi}{6}=\frac{3 \pi}{6}=\frac{\pi}{2}$ and
$
\tan \left(\frac{\pi}{2}\right)=\infty
$

(ii) $\sin \left(\tan ^{-1}\left(\frac{1}{2}\right)-\cos ^{-1}\left(\frac{4}{5}\right)\right)$
Let $\tan ^{-1} \frac{1}{2}=\theta_1 \quad \Rightarrow \tan \theta_1=\frac{1}{2}$
So $\sin \theta_1=\frac{1}{\sqrt{5}}$
Let $\cos ^{-1} \frac{4}{5}=\theta_2$

$
\begin{array}{lr}
\Rightarrow & \cos \theta_2=\frac{4}{5} \\
\Rightarrow & \tan \theta_2=\frac{3}{4}
\end{array}
$
Now $\left(\tan ^{-1} \frac{1}{2}-\cos ^{-1} \frac{4}{5}\right)=\tan ^{-1} \frac{1}{2}-\tan ^{-1} \frac{3}{4}$ $=\tan ^{-1}\left\{\frac{\frac{1}{2}-\frac{3}{4}}{1+\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)}\right\}=\tan ^{-1}\left(\frac{\frac{2-3}{4}}{1+\frac{3}{8}}\right)=\tan ^{-1}\left(\frac{-1}{4} \times \frac{8}{11}\right)$ $=\tan ^{-1}\left(\frac{-2}{11}\right)$

Let $\tan ^{-1}\left(-\frac{2}{11}\right)=\theta_3 \quad \Rightarrow \tan \theta_3=\frac{2}{11}$ $\Rightarrow \quad \sin \theta_3=-\frac{2}{\sqrt{125}} \quad \Rightarrow \theta_3=\sin ^{-1}\left(\frac{-2}{\sqrt{125}}\right)$
So $\sin \left(\tan ^{-1}\left(\frac{1}{2}\right)-\cos ^{-1}\left(\frac{4}{5}\right)\right)=\sin \left[\sin ^{-1}\left(-\frac{2}{\sqrt{125}}\right)\right]=\frac{-2}{\sqrt{125}}$
$
=\frac{-2}{5 \sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=\frac{-2 \sqrt{5}}{25}
$

$\begin{aligned}
& \text { (iii) } \cos \left(\sin ^{-1}\left(\frac{4}{5}\right)-\tan ^{-1}\left(\frac{3}{4}\right)\right) \\
& \text { Let } \sin ^{-1} \frac{4}{5}=\theta_1 \\
& 4 \\
& 5 \\
& \Rightarrow \quad \quad \quad \quad \sin \theta_1=\frac{4}{5} \\
& \Rightarrow \quad \tan \theta_1=\frac{4}{3} \quad \Rightarrow \theta_1=\tan ^{-1} \frac{4}{3} \\
& \text { Now } \sin ^{-1}\left(\frac{4}{5}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=\tan ^{-1} \frac{4}{3}-\tan ^{-1} \frac{3}{4} \\
& =\tan ^{-1}\left\{\frac{\frac{4}{3}-\frac{3}{4}}{1+\left(\frac{4}{3}\right)\left(\frac{3}{4}\right)}\right\}=\tan ^{-1}\left[\frac{\frac{16-9}{12}}{1+\frac{12}{12}}\right]=\tan ^{-1} \frac{7}{24} \\
&
\end{aligned}$

Let

$
\tan ^{-1} \frac{7}{24}=\theta_2 \quad \Rightarrow \tan \theta_2=\frac{7}{24}
$
$\Rightarrow \quad \cos \theta_2=\frac{24}{25}$
$\Rightarrow \quad \theta_2=\cos ^{-1} \frac{24}{25}$
Now $\cos \left[\sin ^{-1}\left(\frac{4}{5}\right)-\tan ^{-1}\left(\frac{3}{4}\right)\right]=\cos \left[\cos ^{-1} \frac{24}{25}\right]=\frac{24}{25}$