Find all positive integral solutions of $\tan^{-1}x+\cos^{-1}\frac{y}{\sqrt{y^2+1}}=\sin^{-1}\frac{3}{\sqrt{10}}$ Find all the positive integral solutions of, $\tan^{-1}x+\cos^{-1}\dfrac{y}{\sqrt{y^2+1}}=\sin^{-1}\dfrac{3}{\sqrt{10}}$
Assuming $x\ge1,y\ge1$ as we have to find positive integral solutions of $(x,y)$
$$\tan^{-1}x=\tan^{-1}3-\tan^{-1}\dfrac{1}{y}$$
As $3>0$ and $\dfrac{1}{y}>0$
$$\tan^{-1}x=\tan^{-1}\left(\dfrac{3-\dfrac{1}{y}}{1+\dfrac{3}{y}}\right)$$
$$\tan^{-1}x=\tan^{-1}\dfrac{3y-1}{y+3}$$
$$x=\dfrac{3y-1}{y+3}$$
$y+3\in[4,\infty)$ as $y\ge1$, $3y-1\in [2,\infty)$ as $y\ge1$
For $x$ to be positive integer, $3y-1$ should be multiple of $y+3$
$$3y-1=m(y+3) \text { where } m\in Z^{+}$$
$$3y-my=3m-1$$
$$(3-m)y=3m-1$$
Here R.H.S is positive, so L.H.S should also be positive.
So $3-m>0$, hence $m<3$
So possible values of $m$ are {$1$,$2$}.
For $m=1$, $$3y-1=y+3$$
$$2y=4$$
$$y=2$$
$$x=\dfrac{3\cdot2-1}{2+3}$$
$$x=1$$
For $m=2$, $$3y-1=2(y+3)$$
$$3y-1=2y+6$$
$$y=7$$
$$x=\dfrac{3\cdot7-1}{7+3}$$
$$x=\dfrac{20}{10}$$
$$x=2$$
Is there some other nicer way to solve this problem.
 A: Your method seems decent. Here I propose another way to ensure the integer-ness of $x$ and $y$.
Note from a rough sketch and limits that the range of $f:\mathbb{R^+} \mapsto X,~f(y)= {3y-1\over y+3}$ is $\left[-\frac 13, 3\right)$. The only positive integers that lie in this range are ${1,2}$. You may use the inverse process $y={3x+1\over 3-x}$ and,use $x\in \{1,2\}$ to see  for which of these $y$, $x$ is also a positive integer. 
For $x=1$, $y=\frac 42 = 2$, which is a positive integer. So one solution is $(x,y)\equiv (1,2)$. For $x=2$, $y=\frac 71 = 7$, also a positive integer. So another solution is $(2,7)$. We have exhausted all possibilities for $y$. Our solution set is $\left\{(1,2),(2,7)\right\}$. 
A: We can actually utilize $x,y$ are positive integers
$$\tan^{-1}x=\tan^{-1}3-\tan^{-1}\dfrac1y<\tan^{-1}3$$
As $\tan^{-1}x$ is an increasing function
$$\implies x<3$$
So, $x$ can be $1$ or $2$
Check which ones make $y$ positive integer
A: Another way
As we need $x,y>0$
If $x/y<1$
$$\tan^{-1}x+\tan^{-1}(1/y)=\tan^{-1}\dfrac{xy+1}{y-x}$$
$$xy+1=3(y-x)$$
$$\iff x=\dfrac{3y-1}{y+3}=3-\dfrac{10}{y+3}$$
So, $y+3(>3), $ must divide $10$ and must honor $x<y$
If $x/y >1,x>y$ as $x,y>0$
$$\tan^{-1}x+\tan^{-1}\dfrac1y>\tan^{-1}y+\cot^{-1}y=\dfrac\pi2>\tan^{-1}3$$
