Exercise 3.11-Additional Questions - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions
Question 1.
Prove that (i) $\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)=\tan ^{-1} \frac{2}{9}$
(ii) $\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{3}{5}=\tan ^{-1} \frac{27}{11}$.
Solution:

(i) $\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{13}}{1-\frac{1}{7} \cdot \frac{1}{13}}\right)=\tan ^{-1}\left(\frac{20}{90}\right)=\tan ^{-1}\left(\frac{2}{9}\right)$
(ii) Let $\cos ^{-1} \frac{4}{5}=\theta$ then $\cos \theta=\frac{4}{5} \therefore \tan \theta=\frac{3}{4} \Rightarrow \theta=\tan ^{-1} \frac{3}{4}$
$
\begin{aligned}
& \cos ^{-1} \frac{4}{5}=\tan ^{-1} \frac{3}{4} \\
\therefore \cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{3}{5}= & \tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5} \\
= & \tan ^{-1}\left\{\frac{\frac{3}{4}+\frac{3}{5}}{1-\left(\frac{3}{4}\right)\left(\frac{3}{5}\right)}\right\}=\tan ^{-1} \frac{27}{11}
\end{aligned}
$

Question 2.
Solve $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$.
Solution:
$
\begin{aligned}
& \tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4} \Rightarrow \tan ^{-1}\left[\frac{2 x+3 x}{1-6 x^2}\right]=\tan ^{-1} \text { (1) } \\
& \qquad \begin{array}{l}
\therefore \frac{5 x}{1-6 x^2}=1 \Rightarrow 1-6 x^2=5 x \Rightarrow 6 x^2+5 x-1=0
\end{array} \\
& \text { (i.e.) }(x+1)(6 x-1)=0 \Rightarrow x=-1 \text { or } \frac{1}{6}
\end{aligned}
$
The negative value of $x$ is rejected since it makes RHS negative $\therefore \mathrm{x}=\frac{1}{6}$
Question 3.
Prove that $\tan ^{-1}\left(\frac{m}{n}\right)-\tan ^{-1}\left(\frac{m-n}{m+n}\right)=\frac{\pi}{4}$

Solution:
$
\begin{aligned}
\text { LHS } & =\tan ^{-1}\left(\frac{m}{n}\right)-\tan ^{-1}\left(\frac{m-n}{m+n}\right) \\
& =\tan ^{-1}\left(\frac{\frac{m}{n}-\frac{m-n}{m+n}}{1+\frac{m}{n} \times \frac{m-n}{m+n}}\right)=\tan ^{-1}\left(\frac{m^2+m n-m n+n^2}{m n+n^2+m^2-m n}\right) \\
& =\tan ^{-1}\left(\frac{m^2+n^2}{m^2+n^2}\right)=\tan ^{-1}(1)=\frac{\pi}{4}
\end{aligned}
$
Question 4.
Solve $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\frac{\pi}{3}$, where $x>0$.
Solution:

$
\begin{gathered}
\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\frac{\pi}{3} \\
\cot ^{-1}\left(\frac{1}{x}\right)=\tan ^{-1} x
\end{gathered}
$
Therefore $\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$
Hence, LHS $=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=\frac{\pi}{3}$
$
\begin{aligned}
\Rightarrow \tan ^{-1} \frac{2 x}{1-x^2} & =\frac{\pi}{6} \\
\frac{2 x}{1-x^2} & =\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \\
1-x^2 & =2 \sqrt{3} x \Rightarrow x^2+2 \sqrt{3} x-1=0 \\
x & =\frac{-2 \sqrt{3} \pm \sqrt{12+4}}{2}=\frac{-2 \sqrt{3} \pm 4}{2} \\
& =2-\sqrt{3} \text { (or) }-2-\sqrt{3} \\
& \therefore x=2-\sqrt{3}
\end{aligned}
$
Question 5.
Solve: $\tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1} \frac{4}{7}, 0$

Solution:
$
\begin{aligned}
\tan ^{-1}(x+1)+\tan ^{-1}(x-1) & =\tan ^{-1}\left(\frac{4}{7}\right) \\
\tan ^{-1}\left(\frac{x+1+x-1}{1-\left(x^2-1\right)}\right) & =\tan ^{-1}\left(\frac{4}{7}\right) \\
\tan \left(\frac{2 x}{2-x^2}\right) & =\tan ^{-1}\left(\frac{4}{7}\right) \\
\frac{2 x}{2-x^2} & =\frac{4}{7} \\
8-4 x^2 & =14 x \\
4 x^2+14 x-8 & =0 \Rightarrow 2 x^2+7 x-4=0 \\
2 x^2+8 x-x-4 & =0 \\
2 x(x+4)-1(x+4) & =0 \\
(x+4)(2 x-1) & =0 \\
x & =-4 \text { or } x=\frac{1}{2} \\
\therefore x=\frac{1}{2} &
\end{aligned}
$