Question

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

पर्याय

•  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(3+y)=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$$