Question

The positive integral solution of the equation
\[\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^2}} = \sin^{- 1} \frac{3}{\sqrt{10}}\text{ is }\]

Options

•  x = 1, y = 2

•  x = 2, y = 1

•  x = 3, y = 2

• x = −2, y = −1

Answer 1

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