Exercise 4.5 - Chapter 4 Inverse Trigonometric Functions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

${Ex} 4.5$
Question 1.
Find the value, if it exists. If not, give the reason for non-existence.
(i) $\sin ^{-1}(\cos \pi)$
(ii) $\tan ^{-1}\left(\sin \left(-\frac{5 \pi}{2}\right)\right)$
(iii) $\sin ^{-1}[\sin 5]$
Solution:
(i) $\sin ^{-1}(\cos \pi)=\sin ^{-1}(-1)=-\frac{\pi}{2}$
(ii) $\tan ^{-1}\left(\sin \frac{5 \pi}{2}\right)=\tan ^{-1}\left(-\sin \frac{\pi}{2}\right)=\tan ^{-1}(-1)=-\frac{\pi}{4}$
(iii) $\sin ^{-1}(\sin 5)=\sin ^{-1}[\sin (5-2 \pi)]=5-2 \pi$

Question 2.
Find the value of the expression in terms of $x$, with the help of a reference triangle.
(i) $\sin \left(\cos ^{-1}(1-\mathrm{x})\right)$
(ii) $\cos \left(\tan ^{-1}(3 x-1)\right)$
(iii) $\tan \left(\sin ^{-1}\left(x+\frac{1}{2}\right)\right)$
Solution:
(i) Let $\cos ^{-1}(1-x)=\theta$
$1-\mathrm{x}=\cos \theta$
We know $\sin ^2 \theta=1-\cos ^2 \theta$

$\begin{aligned}
\Rightarrow \quad \sin \theta & =\sqrt{1-\cos ^2 \theta}=\sqrt{1-(1-x)^2}=\sqrt{1-\left(1+x^2-2 x\right)} \\
& =\sqrt{2 x-x^2} \\
\Rightarrow \quad \sin \left[\cos ^{-1}(1-x)\right] & =\sin \theta=\sqrt{2 x-x^2}
\end{aligned}$

(ii) Let $\tan ^{-1}(3 x-1)=\theta \quad \Rightarrow 3 x-1=\tan \theta$
So $\tan ^2 \theta=(3 x-1)^2=9 x^2-6 x+1$
We know $\quad 1+\tan ^2 \theta=\sec ^2 \theta=\frac{1}{\cos ^2 \theta}$ $\Rightarrow \quad \frac{1}{\cos ^2 \theta}=9 x^2-6 x+1+1=9 x^2-6 x+2$
So $\cos ^2 \theta=\frac{1}{9 x^2-6 x+2} \quad \Rightarrow \cos \theta=\frac{1}{\sqrt{9 x^2-6 x+2}}$ $\therefore \quad \cos \left[\tan ^{-1}(3 x-1)\right]=\frac{1}{\sqrt{9 x^2-6 x+2}}$

$\begin{aligned}
& \text { (iii) Let } \sin ^{-1}\left(x+\frac{1}{2}\right)=\theta \quad \Rightarrow x+\frac{1}{2}=\sin \theta \\
& \Rightarrow \quad 2 x+1=2 \sin \theta \quad \Rightarrow \sin \theta=\frac{2 x+1}{2} \\
& \cos \theta=\sqrt{1-\sin ^2 \theta}=\sqrt{1-\left(\frac{2 x+1}{2}\right)^2} \\
& =\sqrt{1-\frac{\left(4 x^2+4 x+1\right)}{4}}=\frac{1}{2} \sqrt{3-4 x-4 x^2} \\
& \tan (\theta)=\frac{\sin \theta}{\cos \theta}=\frac{\frac{2 x+1}{2}}{\frac{1}{2} \sqrt{3-4 x-4 x^2}}=\frac{2 x+1}{\sqrt{3-4 x-4 x^2}} \\
&
\end{aligned}$

Question 3.
Find the value of
(i) $\sin ^{-1}\left(\cos \left(\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)\right.$
(ii) $\cot \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{4}{5}\right)$
(iii) $\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$

Solution:
(i) $\sin ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{3}$ and $\cos \left(\frac{\pi}{3}\right)=\frac{1}{2}$ and so $\sin ^{-1}\left[\cos \left\{\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}\right]=\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$

$\begin{aligned}
& \text { (ii) } \quad \text { Let } \sin ^{-1} \frac{4}{5}=\theta \\
& \Rightarrow \quad \sin \theta=\frac{4}{5} \\
& \Rightarrow \quad \cos \theta=\frac{3}{5} \quad \Rightarrow \theta=\cos ^{-1} \frac{3}{5} \\
& \therefore \quad \cot \left[\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{4}{5}\right]=\cot \left[\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{3}{5}\right]=\cot \left(\frac{\pi}{2}\right)=0 \\
&
\end{aligned}$

(iii) $\quad$ Let $\sin ^{-1} \frac{3}{5}=\theta$
$
\begin{aligned}
& \Rightarrow \quad \sin \theta=\frac{3}{5} \\
& \Rightarrow \quad \tan \theta=\frac{3}{4} \\
& \text { so } \\
& \theta=\tan ^{-1} \frac{3}{4} \\
& \text { so } \\
& \theta=\tan ^{-1} \frac{3}{4} \\
& \text { Now let } \cot ^{-1} \frac{3}{2}=\theta \quad \Rightarrow \cot \theta=\frac{3}{2} \\
& \Rightarrow \quad \tan \theta=\frac{2}{3} \quad \Rightarrow \theta=\tan ^{-1} \frac{2}{3} \\
& \text { Now } \sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}=\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}}\right) \\
& =\tan ^{-1}\left(\frac{\frac{9+8}{12}}{1-\frac{6}{12}}\right)=\tan ^{-1}\left(\frac{17}{6}\right) \\
& \text { so } \tan \left[\tan ^{-1}\left(\frac{17}{6}\right)\right]=\frac{17}{6} \\
&
\end{aligned}
$

Question 4.
Prove that
(i) $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$
(ii) $\sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{12}{13}=\sin ^{-1} \frac{16}{65}$
Solution:
(i)
$
\begin{aligned}
\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1}\left[\frac{\left(\frac{2}{11}\right)+\left(\frac{7}{24}\right)}{1-\frac{2}{11} \times \frac{7}{24}}\right]=\tan ^{-1}\left[\frac{\frac{48+77}{11 \times 24}}{1-\frac{14}{11 \times 24}}\right] \\
=\tan ^{-1}\left[\frac{\frac{264}{250}}{264}\right]=\tan ^{-1}\left(\frac{125}{250}\right)=\tan ^{-1}\left(\frac{1}{2}\right)
\end{aligned}
$

(ii) $\quad$ Let $\cos ^{-1} \frac{12}{13}=\theta \quad \Rightarrow \cos \theta=\frac{12}{13}$
so
$
\sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}
$
$\Rightarrow \quad \theta=\sin ^{-1} \frac{5}{13}$
Now to find $\sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{5}{13}$
We know $\quad \sin ^{-1} x-\sin ^{-1} y=\sin ^{-1}\left[x \sqrt{1-y^2}-y \sqrt{1-x^2}\right]$
LHS: $\quad \sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{5}{13}=\sin ^{-1}\left[\frac{3}{5} \sqrt{1-\frac{25}{169}}-\frac{5}{13} \sqrt{1-\frac{9}{25}}\right]=\sin ^{-1}\left[\frac{3}{5} \times \frac{12}{13}-\frac{5}{13} \times \frac{4}{5}\right]$ $=\sin ^{-1}\left(\frac{36}{65}-\frac{20}{65}\right)=\sin ^{-1} \frac{16}{65}=$ RHS.
Question 5.
Prove that $\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}+\tan ^{-1} \mathrm{z}=\tan ^{-1}\left[\frac{x+y+z-x y z}{1-x y-y z-z x}\right]$

Solution:
$
\begin{aligned}
\tan ^{-1} x+\tan ^{-1} y & =\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\tan ^{-1}(\mathrm{~A}) \\
\text { Here } \mathrm{A} & =\frac{x+y}{1-x y}
\end{aligned}
$
So LHS: $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\tan ^{-1}$ (A) $+\tan ^{-1} z$
$
\begin{aligned}
\tan ^{-1}\left(\frac{\mathrm{A}+z}{1-\mathrm{A} z}\right) & =\tan ^{-1}\left[\frac{\frac{x+y}{1-x y}+z}{1-\frac{x+y}{1-x y}(z)}\right]=\tan ^{-1}\left[\frac{\frac{x+y+z(1-x y)}{1-x y}}{\frac{1-x y-(x+y) z}{1-x y}}\right] \\
& =\tan ^{-1}\left(\frac{x+y+z-x y z}{1-x y-x z-y z}\right)=\text { RHS }
\end{aligned}
$

Question 6.
If $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi$, show that $x+y+z=x y z$.
Solution:
$
\begin{aligned}
& \text { Given } \tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}+\tan ^{-1} \mathrm{z}=\pi \\
& \Rightarrow \quad \tan ^{-1}\left(\frac{x+y+z-x y z}{1-x y-y z-z x}\right)=\pi \\
& \Rightarrow \quad \frac{x+y+z-x y z}{1-x y-y z-z x}=\tan \pi=0 \\
& \Rightarrow \quad x+y+z-x y z=0 \quad \Rightarrow x+y+z=x y z \\
&
\end{aligned}
$

Question 7.
Prove that $\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^2}=\tan ^{-1} \frac{3 x-x^3}{1-3 x^2},|x|<\frac{1}{\sqrt{3}}$
Solution:
$
\begin{aligned}
& \text { LHS: } \tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^2}=\tan ^{-1}\left[\frac{x+\frac{2 x}{1-x^2}}{1-x\left(\frac{2 x}{1-x^2}\right)}\right]=\tan ^{-1}\left[\frac{\frac{x\left(1-x^2\right)+2 x}{1-x^2}}{1-\frac{x(2 x)}{1-x^2}}\right] \\
& =\tan ^{-1}\left[\frac{\frac{x-x^3+2 x}{1-x^2}}{\frac{1-x^2-2 x^2}{1-x^2}}\right]=\tan ^{-1} \frac{3 x-x^3}{1-3 x^2} \\
& =\mathrm{RHS} \\
&
\end{aligned}
$

Question 8.
Simplify: $\tan ^{-1} \frac{x}{y}-\tan ^{-1} \frac{x-y}{x+y}$
Solution:
$
\begin{aligned}
\tan ^{-1} \frac{x}{y}-\tan ^{-1} \frac{x-y}{x+y} & =\tan ^{-1}\left[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\left(\frac{x}{y}\right)\left(\frac{x-y}{x+y}\right)}\right]=\tan ^{-1}\left[\frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{1+\frac{x(x-y)}{y(x+y)}}\right] \\
& =\tan ^{-1}\left[\frac{\frac{x^2+x-x+y+y^2}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}\right]=\tan ^{-1}\left(\frac{x^2+y^2}{x y+y^2+x^2-x y}\right) \\
& =\tan ^{-1}\left(\frac{x^2+y^2}{x^2+y^2}\right)=\tan ^{-1}(1)=\frac{\pi}{4}
\end{aligned}
$

Question 9.
Find the value of
(i) $\sin ^{-1} \frac{5}{x}+\sin ^{-1} \frac{12}{x}=\frac{\pi}{2}$
(ii) $2 \tan ^{-1} x=\cos ^{-1} \frac{1-a^2}{1+a^2}-\cos ^{-1} \frac{1-b^2}{1+b^2}, \mathrm{a}>0, \mathrm{~b}>0$
(iii) $2 \tan ^{-1}(\cos x)=\operatorname{tarn}^{-1}(2 \operatorname{cosec} x)$
(iv) $\cot ^{-1} x-\cot ^{-1}(x+2)=\frac{\pi}{12}, x>0$
Solution:

$\begin{aligned}
& \text { (i) Given } \sin ^{-1} \frac{5}{x}+\sin ^{-1} \frac{12}{x}=\frac{\pi}{2} \\
& \Rightarrow \quad \sin ^{-1} \frac{5}{x}=\frac{\pi}{2}-\sin ^{-1} \frac{12}{x}=\cos ^{-1} \frac{12}{x} \quad\left(\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right) \\
& \text { so now } \\
& \sin ^{-1} \frac{5}{x}=\cos ^{-1} \frac{12}{x} \\
& \text { Let } \sin ^{-1} \frac{5}{x}=\theta \\
& \text { Let } \cos ^{-1} \frac{12}{x}=\theta \\
& \Rightarrow \quad \sin \theta=\frac{5}{x} \quad \Rightarrow \quad \frac{12}{x}=\cos \theta \\
& \text { But we know } \sin ^2 \theta+\cos ^2 \theta=1 \\
& \Rightarrow \quad\left(\frac{5}{x}\right)^2+\left(\frac{12}{x}\right)^2=1 \\
& \text { i.e. } \quad \frac{25}{x^2}+\frac{144}{x^2}=1 \quad \Rightarrow x^2=169 \\
& \text { a } \\
& x=13 \\
&
\end{aligned}$

(ii)
Let $2 \tan ^{-1} x=\cos ^{-1} \frac{1-a^2}{1+a^2}-\cos ^{-1} \frac{1-b^2}{1+b^2}$
Put $a=\tan \theta_1 \quad \Rightarrow \theta_1=\tan ^{-1} a$
Now $\frac{1-a^2}{1+a^2}=\frac{1-\tan ^2 \theta_1}{1+\tan ^2 \theta_1}=\cos 2 \theta_1$
Similarly, put $b=\tan \theta_2 \quad \Rightarrow \theta_2=\tan ^{-1} b$
Now $\frac{1-b^2}{1+b^2}=\cos 2 \theta_1$
and
$
\text { so } \begin{aligned}
\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right) & =\cos ^{-1}\left(\cos 2 \theta_1\right)=2 \theta_1=2 \tan ^{-1} a \\
\cos ^{-1}\left(\frac{1-b^2}{1+b^2}\right) & =\cos ^{-1}\left(\cos 2 \theta_2\right)=2 \theta_2=2 \tan ^{-1} b
\end{aligned}
$
Now $2 \tan ^{-1} x=2 \tan ^{-1} a-2 \tan ^{-1} b$
$\Rightarrow(\div$ by 2$) \quad \tan ^{-1} x=\tan ^{-1} a-\tan ^{-1} b$
$\tan ^{-1} x=\tan ^{-1}\left(\frac{a-b}{1+a b}\right) \Rightarrow x=\frac{a-b}{1+a b}$

(iii)
$
\begin{aligned}
2 \tan ^{-1}(\cos x) & =\tan ^{-1}(2 \operatorname{cosec} x) \\
\text { LHS } & =2 \tan ^{-1}(\cos x)=\tan ^{-1}(\cos x)+\tan ^{-1}(\cos x) \\
& =\tan ^{-1}\left(\frac{\cos x+\cos x}{1-\cos ^2 x}\right)=\tan ^{-1}\left(\frac{2 \cos x}{\sin ^2 x}\right)
\end{aligned}
$
$
\begin{aligned}
\mathrm{RHS} & =\tan ^{-1}(2 \operatorname{cosec} x) \\
\text { Now LHS } & =\text { RHS } \\
\Rightarrow \quad \tan ^{-1}\left(\frac{2 \cos x}{\sin ^2 x}\right) & =\tan ^{-1}(2 \operatorname{cosec} x) \\
\Rightarrow \quad \frac{2 \cos x}{\sin ^2 x} & =2 \operatorname{cosec} x=\frac{2}{\sin x} \\
2 \cot x & =2 \\
\Rightarrow \quad \cot x & =1 \quad \Rightarrow x=\frac{\pi}{4}
\end{aligned}
$

$\begin{aligned}
& \text { (iv) } \tan ^{-1} \frac{1}{x}-\tan ^{-1} \frac{1}{x+2}=\frac{\pi}{12} \\
& =\tan ^{-1}\left[\frac{\frac{1}{x}-\frac{1}{x+2}}{1+\left(\frac{1}{x}\right)\left(\frac{1}{x+2}\right)}\right]=\tan ^{-1}\left[\frac{\frac{x+2-x}{x(x+2)}}{\frac{x(x+2)+1}{x(x+2)}}\right] \\
& =\tan ^{-1} \frac{2}{x^2+2 x+1}=\frac{\pi}{12} \\
& \Rightarrow \quad \frac{2}{x^2+2 x+1}=\tan \frac{\pi}{12}=\tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right) \\
& =\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}} \\
& \Rightarrow \quad \frac{2}{x^2+2 x+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{3-1}{(\sqrt{3}+1)^2} \\
& \Rightarrow \quad \frac{2}{x^2+2 x+1}=\frac{2}{(\sqrt{3}+1)^2} \\
& \Rightarrow \quad(x+1)^2=(\sqrt{3}+1)^2 \Rightarrow x=\sqrt{3} \\
&
\end{aligned}$

Question 10.
Find the number of solution of the equation $\tan ^{-1}(x-1)+\tan ^{-1} x+\tan ^{-1}(x+1)=\tan ^{-1}(3 x)$.

Solution:
$
\begin{aligned}
& \tan ^{-1}(x-1)+\tan ^{-1} x+\tan ^{-1}(x+1)=\tan ^{-1}(3 x) \\
& \tan ^{-1}(x-1)+\tan ^{-1}(x+1)=\tan ^{-1} 3 x-\tan ^{-1} x
\end{aligned}
$
$
\begin{aligned}
& \text { LHS: } \quad \tan ^{-1} \frac{(x-1)+(x+1)}{1-(x-1)(x+1)}=\tan ^{-1} \frac{2 x}{1-\left(x^2-1\right)}=\tan ^{-1} \frac{2 x}{2-x^2} \\
& \text { RHS: } \quad \tan ^{-1} 3 x-\tan ^{-1} x=\tan ^{-1} \frac{3 x-x}{1+(3 x)(x)}=\tan ^{-1} \frac{2 x}{1+3 x^2} \\
& \text { LHS }=\text { RHS } \\
& \Rightarrow \tan ^{-1} \frac{2 x}{2-x^2}=\tan ^{-1} \frac{2 x}{1+3 x^2} \\
& \Rightarrow \frac{2 x}{2-x^2}=\frac{2 x}{1+3 x^2} \\
& \Rightarrow 2-\mathrm{x}^2=1+3 \mathrm{x}^2 \\
& \Rightarrow 4 \mathrm{x}^2=1 \\
& \Rightarrow \mathrm{x}^2=\frac{1}{4} \\
& \Rightarrow \mathrm{x}=\pm \frac{1}{2}
\end{aligned}
$
So, the equation has 2 solutions.