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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
Find the principle value of:

$
\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)
$
Solution:
Let $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=y$ then, $\frac{1}{\sqrt{3}}=\tan y$
This gives
$
y=\frac{\pi}{6} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
$
$
\therefore \quad \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}
$
Question 2.
Find the value of

$
\tan ^{-1}\left(\tan \frac{4 \pi}{3}\right)
$
Solution:
$
\tan ^{-1}\left(\tan \frac{4 \pi}{3}\right)=\tan ^{-1}\left(\tan \left(\pi+\frac{\pi}{3}\right)\right)=\tan ^{-1}\left(\tan \frac{\pi}{3}\right)=\frac{\pi}{3}
$
Since
$
\frac{\pi}{3} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
$

Question 3.
Find the value of $\sin ^2\left(\tan ^{-1} \frac{3}{4}\right)$
Solution:
$
\text { Consider } \begin{aligned}
\tan ^{-1} \frac{3}{4} & =\theta \\
\frac{3}{4} & =\tan \theta \\
\frac{3}{5} & =\sin \theta \\
\sin ^2\left(\tan ^{-1} \frac{3}{4}\right) & =\left[\sin \left(\tan ^{-1} \frac{3}{4}\right)\right]^2=\left[\sin \left(\sin ^{-1}\left(\frac{3}{5}\right)\right)\right]^2=\left(\frac{3}{5}\right)^2=\frac{9}{25}
\end{aligned}
$