The positive integral solution of the equation

Question:

The positive integral solution of the equation

$\tan ^{-1} x+\cos ^{-1} \frac{y}{\sqrt{1+y^{2}}}=\sin ^{-1} \frac{3}{\sqrt{10}} \mathrm{i}$

(a) $x=1, y=2$

(b) $x=2, y=1$

(c) $x=3, y=2$

(d) $x=-2, y=-1$.

Solution:

(a) $x=1, y=2$

We have,

$\tan ^{-1} x+\cos ^{-1} \frac{y}{\sqrt{1+y^{2}}}=\sin ^{-1} \frac{3}{\sqrt{10}}$

$\Rightarrow \tan ^{-1} x+\tan ^{-1}\left[\frac{\sqrt{\sqrt{1-\left(\frac{y}{\sqrt{1+y^{2}}}\right)^{2}}}}{\frac{y}{\sqrt{1+y^{2}}}}\right]=\tan ^{-1}\left[\frac{\frac{3}{\sqrt{N}}}{\sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^{2}}}\right]$

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

$\Rightarrow \tan ^{-1}\left(\frac{x+\frac{1}{y}}{1-x \times \frac{1}{y}}\right)=\tan ^{-1}(3)$

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

$\Rightarrow 3 y-3 x=x y+1$

$\Rightarrow 3 x+x y=3 y-1$

$\Rightarrow x(3+y)=3 y-1$

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

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, $y=1 \Rightarrow x=\frac{7}{3}$ and so on......

Therefore, only integral solutions are :

$x=1$ and $y=2$