Exercise 3.11 - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 3.11
Question 1.
Find the principal value of $\left(\right.$ (i) $\sin ^{-1} \frac{1}{\sqrt{2}}$

(ii) $\cos ^{-1} \frac{\sqrt{3}}{2}$

(iii) $\operatorname{cosec}^{-1}(-1)$

(iv) $\sec ^{-1}(-\sqrt{2})$
(v) $\tan ^{-1}(\sqrt{3})$
Solution:
(i) $\sin ^{-1} \frac{1}{\sqrt{2}}=\theta$
So $\sin \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}$
(ii) $\cos ^{-1} \frac{\sqrt{3}}{2} \quad=\theta \Rightarrow \cos \theta=\frac{\sqrt{3}}{2}=\frac{\pi}{6}$
(iii) $\operatorname{cosec}^{-1}(-1) \quad=\theta \Rightarrow \operatorname{cosec} \theta-1$
(i.e.) $\frac{1}{\sin \theta}=-1 \Rightarrow \sin \theta=-1=-\frac{\pi}{2}$
(iv) $\sec ^{-1}(-\sqrt{2}) \quad=\theta \Rightarrow \sec \theta=-\sqrt{2}$
$
\begin{aligned}
& \Rightarrow \cos \theta=-\frac{1}{\sqrt{2}} \\
& \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}
\end{aligned}
$
(v) $\tan ^{-1}(\sqrt{3}) \quad=\theta \Rightarrow \tan \theta=\sqrt{3} \Rightarrow \theta=\frac{\pi}{3}$
Question 2.
A man standing directly opposite to one side of a road of width $\mathrm{x}$ meter views a circular shaped traffic green signal of diameter a meter on the other side of the road. The bottom of the green signal is $\mathrm{b}$ meter height from the horizontal level of viewer's eye. If a denotes the angle subtended by the diameter of the green signal at the viewer's eye, then prove that

$
\alpha=\tan ^{-1}\left(\frac{a+b}{x}\right)-\tan ^{-1}\left(\frac{b}{x}\right)
$
Solution

From the right angled triangle $\mathrm{ABC}$
$
\tan \theta=\frac{b}{x} \Rightarrow \theta=\tan ^{-1}\left(\frac{b}{x}\right)
$
From the right angled triangle $\mathrm{ABD}$
$
\begin{aligned}
& \begin{aligned}
\tan (\theta+\alpha)= & \frac{a+b}{x} \Rightarrow \theta+\alpha=\tan ^{-1} \frac{a+b}{x} \\
\text { So } \alpha & =\tan ^{-1}\left(\frac{a+b}{x}\right)-\theta \\
& =\tan ^{-1}\left(\frac{a+b}{x}\right)-\tan ^{-1}\left(\frac{b}{x}\right)
\end{aligned} \\
& \text { From (1) } \alpha=\tan ^{-1}\left(\frac{a+b}{x}\right)-\tan ^{-1}\left(\frac{b}{x}\right)
\end{aligned}
$