Exercise 4.5-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.
Solve the following equation: $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$
Solution:
$
\begin{aligned}
& \text { If } \sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2} \\
& \Rightarrow \quad \sin ^{-1}(1-x)=\frac{\pi}{2}+2 \sin ^{-1} x \\
& \Rightarrow \quad 1-x=\sin \left[\frac{\pi}{2}+2 \sin ^{-1} x\right]=\cos \left(2 \sin ^{-1} x\right) \\
& \Rightarrow \quad 1-x=\cos \left[\cos ^{-1}\left(1-2 x^2\right)\right] \quad\left\{\because 2 \sin ^{-1} x=\cos ^{-1}\left(1-2 x^2\right)\right\} \\
& \Rightarrow \quad 1-x=1-2 x^2 \quad \Rightarrow 2 x^2-x=0 \\
& \Rightarrow \quad x(2 x-1)=0 \quad \Rightarrow x=0 \text { or } \frac{1}{2} \\
&
\end{aligned}
$
Question 2.
Solve: $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$

Solution:
$
\begin{aligned}
& \text { Now } \quad \tan ^{-1} 2 x+\tan ^{-1} 3 x=\tan ^{-1} \frac{2 x+3 x}{1-(2 x)(3 x)}=\tan ^{-1} \frac{5 x}{1-6 x^2}=\frac{\pi}{4} \text { (Given) } \\
& \Rightarrow \quad \frac{5 x}{1-6 x^2}=\tan \left(\frac{\pi}{4}\right)=1\left(\text { when } 1-6 x^2>0\right) \\
& \Rightarrow \quad 1-6 x^2=5 x \quad \Rightarrow 6 x^2+5 x-1=0 \\
& \Rightarrow \quad(6 x-1)(x+1)=0 \quad \Rightarrow x=\frac{1}{6} \\
&
\end{aligned}
$
Question 3.
Prove: $\sin ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1} \frac{63}{16}$
Solution:
Do it yourself
Question 4.
Prove: $\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)=\tan ^{-1} \frac{2}{9}$
Solution:
Do it yourself
Question 5.
Prove: $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$
Solution:
Do it yourself
Question 6.
Solve for $x: \tan \left(\cos ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{1}{2}\right)$
Solution:
Do it yourself
Question 7.
Solve for $x: \tan \left(\cos ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{1}{2}\right)$

Solution:
$
\begin{aligned}
\tan \left(\cos ^{-1} x\right) & =\tan \left[\tan ^{-1} \frac{\sqrt{1-x^2}}{x}\right] \\
\sin \left(\cot ^{-1} \frac{1}{2}\right) & =\sin \left(\sin ^{-1} \frac{2}{\sqrt{5}}\right) \\
\Rightarrow \quad \frac{\sqrt{1-x^2}}{x} & =\frac{2}{\sqrt{5}} \quad \Rightarrow x=\pm \frac{\sqrt{5}}{3}
\end{aligned}
$
Question 8.
If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$, then find $x$.
Solution:
$
\begin{aligned}
& \sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1 \\
& \Rightarrow \quad \sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x=\sin ^{-1} 1=\frac{\pi}{2} \\
& \Rightarrow \quad x=\frac{1}{5} \\
& \left(\because \sin ^{-1} \frac{1}{5}+\cos ^{-1} \frac{1}{5}=\frac{\pi}{2}\right) \\
&
\end{aligned}
$
Question 9.
If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\pi$, then prove that $x^4+y^4+z^4+4 x^2 y^2 z^2=2\left(x^2 y^2+y^2 z^2+z^2 x^2\right)$.

Solution:

Let $\sin ^{-1} x=\alpha ; \sin ^{-1} y=\beta$ and $\sin ^{-1} z=\gamma$ then $\sin \alpha=x ; \sin \beta=y$ and $\sin \gamma=z$
$
\therefore \begin{aligned}
\alpha+\beta+\gamma & =\pi \\
\alpha+\beta & =\pi-\gamma \\
\cos \alpha \cos \beta-\sin \alpha \sin \beta & =-\cos \gamma \\
\sqrt{1-x^2} \sqrt{1-y^2}-x y & =-\sqrt{1-z^2} \\
\sqrt{1-x^2} \sqrt{1-y^2} & =x y-\sqrt{1-z^2}
\end{aligned} \quad \Rightarrow \cos (\alpha+\beta)=\cos (\pi-\gamma)
$
Squaring on both sides, we get
$
\begin{gathered}
\left(1-x^2\right)\left(1-y^2\right)=x^2 y^2+1-z^2-2 x y \sqrt{1-z^2} \\
1-y^2-x^2+x^2 y^2=x^2 y^2+1-z^2-2 x y \sqrt{1-z^2} \\
2 x y \sqrt{1-z^2}=x^2+y^2-z^2
\end{gathered}
$
Again squaring on both sides, we get
$
\begin{aligned}
4 x^2 y^2\left(1-z^2\right) & =\left(x^2+y^2-z^2\right)^2 \\
4 x^2 y^2-4 x^2 y^2 z^2 & =x^4+y^4+z^4+2 x^2 y^2-2 y^2 z^2-2 x^2 z^2 \\
x^4+y^4+z^4+4 x^2 y^2 z^2 & =2\left(x^2 y^2+y^2 z^2+x^2 z^2\right)
\end{aligned}
$
Question 10.
Prove that: $\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]=\sqrt{\frac{x^2+1}{x^2+2}}$

Solution:
Let $\theta=\cot ^{-1} x$
$
\cot \theta=x ; \theta \in(0, \pi)
$
$
\sin \left(\cot ^{-1} x\right)=\sin \theta=\frac{1}{\operatorname{cosec} \theta}=\frac{1}{\sqrt{1+\cot ^2 \theta}}=\frac{1}{\sqrt{1+x^2}}
$
Now
$
\begin{aligned}
\tan ^{-1}\left[\sin \left(\cot ^{-1} x\right)\right] & =\tan ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \\
\text { Take } \alpha & =\tan ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \\
\tan \alpha & =\frac{1}{\sqrt{1+x^2}} \operatorname{since} \alpha \in\left(0, \frac{\pi}{2}\right) \\
\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right] & =\cos \alpha=\frac{1}{\sec \alpha} \\
& =\frac{1}{\sqrt{1+\tan ^2 \alpha}}=\frac{1}{\sqrt{1+\frac{1}{1+x^2}}}=\sqrt{\frac{2+x^2}{1+x^2}}=\sqrt{\frac{1+x^2}{2+x^2}}
\end{aligned}
$