Exercise 4.4-Additional Problems - Chapter 4 - Inverse Trigonometric Functions - 12th Maths Guide Samacheer Kalvi Solutions

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Additional Problems
Question 1.
Find the principal value of the following
(i) $\cot ^{-1}(-1)$
(ii) $\sec ^{-1}(-\sqrt{2})$
(iii) $\operatorname{cosec}^{-1}\left(\frac{2}{\sqrt{3}}\right)$
Solution:

$
\begin{aligned}
\text { (i) Consider } & \cot \frac{3 \pi}{4} & =\cot \left(\pi-\frac{\pi}{4}\right)=-\cot \frac{\pi}{4} \\
\therefore & \cot \left(\frac{3 \pi}{4}\right) & =-1
\end{aligned}
$
Since $\frac{3 \pi}{4} \in(0, \pi)$, so $\cot ^{-1}(-1)=\frac{3 \pi}{4}$

(ii) Consider
$
\begin{aligned}
& \sec \frac{3 \pi}{4}=\sec \left(\pi-\frac{\pi}{4}\right)=-\sec \frac{\pi}{4} \\
& \sec \frac{3 \pi}{4}=-\sqrt{2}
\end{aligned}
$
Since $\frac{3 \pi}{4} \in\left(-\frac{\pi}{2}, \pi\right)$ so, $\sec ^{-1}(-\sqrt{2})=\frac{3 \pi}{4}$

(iii) Consider $\quad \operatorname{cosec} \frac{\pi}{3}=\frac{2}{\sqrt{3}}$
Since $\frac{\pi}{3} \in\left(0, \frac{\pi}{2}\right]$ so $\cos ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{3}$

Question 2.
Find the value of $\sec ^2\left(\cot ^{-1} 3\right)+\operatorname{cosec}^2\left(\tan ^{-1} 2\right)$

Solution:
$
\begin{aligned}
& \text { If } \theta=\cot ^{-1}(3) \\
& \cot \theta=3 \text { and } \tan \theta=\frac{1}{3} \\
& \therefore \quad \sec ^2 \theta=1+\tan ^2 \theta=1+\left(\frac{1}{3}\right)^2 \quad \Rightarrow \sec ^2 \theta=\frac{10}{9} \\
& \therefore \quad \sec ^2\left(\cot ^{-1} 3\right)=\sec ^2 \theta=\frac{10}{9} \\
& \text { Again, if } \quad \alpha=\tan ^{-1} \text { (2) } \\
& \tan \alpha=2 \text { and } \cot \alpha=\frac{1}{2} \\
& \therefore \quad \operatorname{cosec}^2 \alpha=1+\cot ^2 \alpha=1+\left(\frac{1}{2}\right)^2=\frac{5}{4} \\
& \therefore \quad \operatorname{cosec}^2\left(\tan ^{-1} 2\right)=\operatorname{cosec}^2 \alpha=\frac{5}{4} \\
& \text { Hence } \sec ^2\left(\cot ^{-1} 3\right)+\operatorname{cosec}^2\left(\tan ^{-1} 2\right)=\frac{10}{9}+\frac{5}{4}=\frac{85}{36} \\
&
\end{aligned}
$

Question 3.
Find the value of $
\tan \left(2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right)
$
Solution:
Let $\tan ^{-1}\left(\frac{1}{5}\right)=\theta \quad \Rightarrow \tan \theta=\frac{1}{5} \quad \Rightarrow \theta \in\left(0, \frac{\pi}{2}\right)$
But $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}=\frac{2\left(\frac{1}{5}\right)}{1-\left(\frac{1}{5}\right)^2}=\frac{\frac{2}{5}}{1-\frac{1}{25}}=\frac{2}{5} \times \frac{25}{24}=\frac{5}{12}$
Now $\tan \left(2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right)=\tan \left(2 \alpha-\frac{\pi}{4}\right)=\frac{\tan 2 \alpha-\tan \frac{\pi}{4}}{1+\tan 2 \alpha \tan \frac{\pi}{4}}=\frac{\frac{5}{12}-1}{1+\left(\frac{5}{12}\right)(1)}=\frac{\frac{-7}{12}}{\frac{17}{12}}=\frac{-7}{17}$