Question

The number of the positive integral solutions of ${\tan }^{-1}x+{\cos }^{-1}\left(\frac{y}{\sqrt{(1+y^2)}}\right)={\sin }^{-1}\left(\frac{3}{\sqrt{10}}\right)$ is

• A. $1$
• B. $3$
• C. $2$
• D. $4$

Answer 1

The correct option is A $1$

$\begin{array}{c}\text{Solution}\to {\tan }^{-1}x+{\cos }^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)={\sin }^{-1}\left(\frac{3}{\sqrt{10}}\right)\\ \text{for}{\cos }^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)\\ \text{Let}\theta ={\cos }^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)\\ \Rightarrow \cos \theta =\frac{y}{\sqrt{1+y^2}}\\ \text{Using pythagoras Theorem}\\ \tan \theta =\frac{1}{y}\Rightarrow \theta ={\tan }^{-1}\frac{1}{y}\end{array}$

$\begin{array}{c}\text{Now, let}\alpha ={\sin }^{-1}\left(\frac{3}{\sqrt{10}}\right)\\ \Rightarrow \sin \alpha =\frac{3}{\sqrt{10}}\\ \text{Using Pythagoras Therem}\\ \begin{array}{cc}\mathrm{Tan}\alpha =\frac{3}{1}& \Rightarrow \alpha ={\tan }^{-1}\left(3\right)\\ \Rightarrow {\tan }^{-1}x+{\cos }^{+}\left(\frac{y}{\sqrt{1+y^2}}\right)& ={\sin }^{-1}\left(\frac{3}{\sqrt{10}}\right)\end{array}\end{array}$

$\Rightarrow {\tan }^{-1}x+{\tan }^{-1}\frac{1}{y}={\tan }^{-1}3$

$\begin{array}{c}using{\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)\\ \Rightarrow {\tan }^{-1}\left(\frac{x+\frac{1}{y}}{1-xy}\right)={\tan }^{-1}3\end{array}$

$\Rightarrow \frac{x+1/y}{1-\left(x\right)\left(1/y\right)}=3$ $\Rightarrow yx+1=3y-3x$

on solving the equation we get $x=1;y=2$

Hence, (A) is the correct option.