Exercise 4.4 - 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 $4.4$
Question 1.
Find the principal value of
(i) $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
(ii) $\cot ^{-1}(\sqrt{3})$
(iii) $\operatorname{cosec}^{-1}(-\sqrt{ } 2)$
Solution:
(i) Let $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\theta$
$\Rightarrow \sec \theta=\frac{2}{\sqrt{3}}$
$\Rightarrow \cos \theta=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6}$
$\Rightarrow \theta=\frac{\pi}{6}$
(ii) Let $\cot ^{-1}(\sqrt{ } 3)=\theta$
$\Rightarrow \cot \theta=\sqrt{ } 3$
$\Rightarrow \tan \theta=\frac{1}{\sqrt{3}}=\tan \frac{\pi}{6}$
$\Rightarrow \theta=\frac{\pi}{6}$
(iii) Let $\operatorname{cosec}^{-1}(-\sqrt{2})=\theta$
$\Rightarrow \operatorname{cosec} \theta=-\sqrt{ } 2$
$\Rightarrow \sin \theta=-\frac{1}{\sqrt{2}}$
$\Rightarrow \theta=-\frac{\pi}{4}$
Question 2.
Find the value of
(i) $\tan ^{-1}(\sqrt{ } 3)-\sec ^{-1}(-2)$
(ii) $\sin ^{-1}(-1)+\cos ^{-1}\left(\frac{1}{2}\right)+\cot ^{-1}(2)$
(iii) $\cot ^{-1}(1)+\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)-\sec ^{-1}(-\sqrt{2})$
Solution:

$\begin{aligned}
& \text { (i) Let } \quad \tan ^{-1} \sqrt{3}=\frac{\pi}{3} \\
& \sec ^{-1}(-2)=\theta \quad \Rightarrow \sec \theta=-2 \\
& \Rightarrow \quad \cos \theta=\frac{-1}{2} \quad \Rightarrow \theta=\pi-\frac{\pi}{3}=\frac{2 \pi}{3} \\
& \therefore \tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)=\frac{\pi}{3}-\left(\frac{2 \pi}{3}\right)=-\frac{\pi}{3} \\
&
\end{aligned}$

$\begin{aligned}
& \text { (ii) } \sin ^{-1}(-1)=-\frac{\pi}{2} \\
& \cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \\
& \therefore \sin ^{-1}(-1)+\cos ^{-1}\left(\frac{1}{2}\right)+\cot ^{-1}(2)=-\frac{\pi}{2}+\frac{\pi}{3}+\cot ^{-1}(2)=-\frac{\pi}{6}+\cot ^{-1}(2)
\end{aligned}$

(iii)
$
\begin{aligned}
\cot ^{-1}(1) & =\frac{\pi}{4} \\
\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right) & =-\frac{\pi}{3} \\
\sec ^{-1}(-\sqrt{2}) & =\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}
\end{aligned}
$
So $\cot ^{-1}(1)+\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)-\sec ^{-1}(-\sqrt{2})=\frac{\pi}{4}-\frac{\pi}{3}-\frac{3 \pi}{4}=\frac{-5 \pi}{6}$