Exercise 4.6-Additional Problems - Chapter 4 Inverse Trigonometric Functions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions
Question 1.
Find the principal value of $\sin ^{-1}\left(\frac{1}{2}\right)$ and $\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$.

Solution:

$\frac{\pi}{6}$ and $-\frac{\pi}{4}$
Question 2.

Find the principal value of

$\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$ and $\cos ^{-1}\left(\frac{-1}{2}\right)$.
Solution:

$\frac{5 \pi}{6}$ and $\frac{2 \pi}{3}$
Question 3.
$
\tan ^{-1} 1+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)
$
Solution:
$
\frac{3 \pi}{4}
$
Question 4.
Evaluate
(i) $\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$
(ii) $\tan ^{-1}\left(\tan \frac{\pi}{4}\right)$
Solution:
(i) $-\frac{\pi}{4} ;

$ (ii) $\frac{\pi}{4}$

Question 5.
Evaluate
(i) $\sin ^{-1}\left(\sin \frac{\pi}{3}\right)$
(ii) $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)$
(iii) $\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$
Solution:
(i) $\frac{\pi}{3}$;
(ii) $\frac{2 \pi}{3}$;
(iii) $\frac{5 \pi}{6}$

Question 6.
Simplify: $\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right) \frac{-\pi}{2}-1$
Question 7.
(i) $\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$
(ii) $\tan \left\{\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right\}$
Solution:
(i) $-\frac{7}{17} ;$ (ii) $\frac{3-\sqrt{5}}{2}$
Question 8.
Solve: $\tan ^{-1} \frac{x-1}{x+1}+\tan ^{-1} \frac{2 x-1}{2 x+1}=\tan ^{-1} \frac{23}{26}$
Solution:
$
\mathrm{x}=\frac{1}{6}
$
Question 9.
Find the values of each of the following:
(i) $\tan ^{-1}\left[2 \cos \left\{2 \sin ^{-1}\left(\frac{1}{2}\right)\right\}\right]$
(ii) $\cot \left[\tan ^{-1} a+\cot ^{-1} a\right]$
(iii) $\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right),|x| \geq 1$
Solution:
(i) $\frac{\pi}{4}$; (ii) 0 ; (iii) 0

Question 10.
Solve for $\mathrm{x}$ :
(i) $\tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1} \frac{18}{31}$
(ii) $\sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{3}$
(iii) $\tan \left(\cos ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{1}{2}\right)$
(iv) $\cot ^{-1} x-\cot ^{-1}(x+2)=\frac{\pi}{12}$, where $x>0$
Solution:
(i) $\frac{1}{4} ;$ (ii) $\frac{1}{2} \sqrt{\frac{3}{7}} ;$ (iii) $\pm \frac{\sqrt{5}}{3} ;$ (iv) $\sqrt{3}$
Question 11.
Prove:
(i) $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}$
(ii) $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi$
(iii) $\tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\frac{17}{6}$
Question 12.
Evaluate: $\sin \left(\tan ^{-1} x+\cot ^{-1} x\right)$
Question 13.
The value of $\sin ^{-1}(1)+\sin ^{-1}(0)$ is .......
(a) $\frac{\pi}{2}$
(b) 0
(c) 1
(d) $\pi$
Solution:
(a) $\frac{\pi}{2}$
Hint:
$
\sin ^{-1} 1=\frac{\pi}{2} ; \sin ^{-1} 0=0 \quad \therefore \sin ^{-1} 1+\sin ^{-1} 0=\frac{\pi}{2}+0=\frac{\pi}{2}
$

Question 14.
$
\sin ^{-1}\left(3 \frac{x}{2}\right)+\cos ^{-1}\left(3 \frac{x}{2}\right)=
$
(a) $\frac{3 \pi}{2}$
(b) $6 x$
(c) $3 x$
(d) $\frac{\pi}{2}$
Solution:
(d) $\frac{\pi}{2}$
Hint:
We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
$
\therefore \quad \sin ^{-1}\left(3 \frac{x}{2}\right)+\cos ^{-1}\left(3 \frac{x}{2}\right)=\frac{\pi}{2}
$
Question 15.
$
\tan ^{-1} \mathrm{x}+\cot ^{-1} \mathrm{x}=\text {. }
$
(a) 1
(b) $-\pi$
(c) $\frac{\pi}{2}$
(d) $\pi$
Solution:
(c) $\frac{\pi}{2}$
Hint:
We know that $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$

Question 16.
$
\sin ^{-1} x-\cos ^{-1}(-x)=
$
(a) $\frac{-\pi}{2}$
(b) $\frac{\pi}{2}$
(c) $\frac{-3 \pi}{2}$
(d) $\frac{3 \pi}{2}$
Solution:
(a) $\frac{-\pi}{2}$
Hint:
$
\begin{aligned}
\cos ^{-1}(-x) & =\pi-\cos ^{-1} x \\
\therefore \quad \sin ^{-1} x-\cos ^{-1}(-x) & =\sin ^{-1} x-\left[\pi-\cos ^{-1} x\right]=\sin ^{-1} x-\pi+\cos ^{-1} x \\
& =-\pi+\sin ^{-1} x+\cos ^{-1} x
\end{aligned}
$
But $\quad \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \quad \therefore$ (1) becomes $-\pi+\frac{\pi}{2}=\frac{-\pi}{2}$
Question 17.
$
\sec ^{-1}\left(\frac{2}{3}\right)+\operatorname{cosec}^{-1}\left(\frac{2}{3}\right)=.
$
(a) $\frac{-\pi}{2}$
(b) $\frac{\pi}{2}$
(c) $\pi$
(d) $-\pi$
Solution:
(b) $\frac{\pi}{2}$
Hint:
We know that $\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}$
$
\sec ^{-1}\left(\frac{2}{3}\right)+\operatorname{cosec}^{-1}\left(\frac{2}{3}\right)=\frac{\pi}{2}
$

Question 18.
$
\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)=
$
(a) $\sin ^{-1} \frac{1}{\sqrt{2}}$
(b) $\sin ^{-1}\left(\frac{1}{2}\right)$
(c) $\tan ^{-1}\left(\frac{1}{2}\right)$
(d) $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
Solution:
(a) $\sin ^{-1} \frac{1}{\sqrt{2}}$
Hint:
$
\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1} \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}=\tan ^{-1}\left(\frac{\frac{3+2}{6}}{1-\frac{1}{6}}\right)=\tan ^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)=\tan ^{-1} 1
$
Now $\quad \tan ^{-1} 1=x$ (say) $\quad \therefore 1=\tan x$, i.e., $x=\frac{\pi}{4}$
So $\quad \sin x=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \quad \therefore x=\sin ^{-1} \frac{1}{\sqrt{2}}$
Question 19.
The value of $\cos ^{-1}(-1)+\tan ^{-1}(\infty)+\sin ^{-1} 1=$
(a) $-\pi$
(b) $\frac{3 \pi}{2}$
(c) $30^{\circ}$
(d) $2 \pi$
Solution:
(d) $2 \pi$
Hint:
$
\begin{aligned}
& \cos ^{-1}(-1)=\pi ; \tan ^{-1}(\infty)=\frac{\pi}{2} ; \sin ^{-1}(1)=\frac{\pi}{2} \\
& \therefore \quad \cos ^{-1}(-1)+\tan ^{-1}(\infty)+\sin ^{-1}(1)=\pi+\frac{\pi}{2}+\frac{\pi}{2}=2 \pi \\
&
\end{aligned}
$

Question 20.
The value of $\cos ^{-1}(-1)+\tan ^{-1}(\infty)+\sin ^{-1} 1=$
(a) $\frac{3 \pi}{2}$
(b) $-\pi$
(c) $2 \pi$
(d) $3 \pi$
Solution:
(c) $2 \pi$
Hint:
$
\begin{aligned}
\cos ^{-1}(-1) & =\pi ; \tan ^{-1}(\infty)=\frac{\pi}{2} ; \sin ^{-1}(1)=\frac{\pi}{2} \\
\therefore \quad \cos ^{-1}(-1)+\tan ^{-1}(\infty)+\sin ^{-1}(1) & =\pi+\frac{\pi}{2}+\frac{\pi}{2}=2 \pi
\end{aligned}
$