Question

The positive integral solution of the equation
${\tan }^{-1}x+{\cos }^{-1}\frac{y}{\sqrt{1+y^2}}={\sin }^{-1}\frac{3}{\sqrt{10}}\mathrm{is}$
(a) x = 1, y = 2
(b) x = 2, y = 1
(c) x = 3, y = 2
(d) x = −2, y = −1.

Answer 1

(a) x = 1, y = 2

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {\tan }^{-1}x+{\cos }^{-1}\frac{y}{\sqrt{1+y^2}}={\sin }^{-1}\frac{3}{\sqrt{10}} \\ \Rightarrow {\tan }^{-1}x+{\tan }^{-1}\left[\frac{\sqrt{1-{\left(\frac{y}{\sqrt{1+y^2}}\right)}^2}}{\frac{y}{\sqrt{1+y^2}}}\right]={\tan }^{-1}\left[\frac{{\displaystyle \frac{3}{\sqrt{10}}}}{\sqrt{1-{\left(\frac{3}{\sqrt{10}}\right)}^2}}\right] \\ \Rightarrow {\tan }^{-1}x+{\tan }^{-1}\left(\frac{1}{y}\right)={\tan }^{-1}\left(3\right) \\ \Rightarrow {\tan }^{-1}\left(\frac{x+{\displaystyle \frac{1}{y}}}{1-x\times {\displaystyle \frac{1}{y}}}\right)={\tan }^{-1}\left(3\right) \\ \Rightarrow \frac{xy+1}{y-x}=3 \\ \Rightarrow 3y-3x=xy+1 \\ \Rightarrow 3x+xy=3y-1 \\ \Rightarrow x\left(3+y\right)=3y-1 \\ \Rightarrow x=\frac{3y-1}{3+y} \\ \mathrm{For},y=1\Rightarrow x=\frac{1}{2} \\ \mathrm{For},y=2\Rightarrow x=1 \\ \mathrm{For},y=3\Rightarrow x=\frac{4}{3} \\ \mathrm{For},y=4\Rightarrow x=\frac{11}{7} \\ \mathrm{For},y=1\Rightarrow x=\frac{7}{3}\mathrm{and}\mathrm{so}\mathrm{on}...... \\ \mathrm{Therefore},\mathrm{only}\mathrm{integral}\mathrm{solutions}\mathrm{are}: \\ x=1\mathrm{and}y=2 \end{gathered}$$