Consider the condition xy<1, then find the value of ${{\tan }^{-1}}x+{{\tan }^{-1}}y$

Hint: Use the trigonometric identity related to tan that is $\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$ . Assume ${{\tan }^{-1}}x=A$, similarly with other values and apply the subsequent formula.

Let ${{\tan }^{-1}}x=A$
Multiplying with ‘tan’ on both sides, we get
$tan\left( {{\tan }^{-1}}x \right)=\tan A$
We know $\tan ,{{\tan }^{-1}}$ gets cancelled, so we get
$x=\tan A.........(i)$
Now similarly, let ${{\tan }^{-1}}y=B$
Multiplying with ‘tan’ on both sides, we get
$tan\left( {{\tan }^{-1}}y \right)=\tan B$
We know $\tan ,{{\tan }^{-1}}$ gets cancelled, so we get
$y=\tan B.........(ii)$
Now we know, as per the trigonometric identity, $\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$, so we can write,
$\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
Now substituting the values from equation (i) and (ii), we get
$\tan (A+B)=\dfrac{x+y}{1-xy}$
Now for this to be true, we know that denominator should be greater than 1, so we can write as
1 – xy > 0
Adding ‘xy’ on both sides, we get
1 – xy + xy > xy
Cancelling the like terms, we get
1 > xy
So, $\tan (A+B)=\dfrac{x+y}{1-xy}$ is true only when xy < 1.
Now we will multiply the above expression with ${{\tan }^{-1}}$ , we get
$ta{{n}^{-1}}\left( \tan (A+B) \right)=ta{{n}^{-1}}\left( \dfrac{x+y}{1-xy} \right)$
We know $\tan ,{{\tan }^{-1}}$ gets cancelled, so we get
$A+B=ta{{n}^{-1}}\left( \dfrac{x+y}{1-xy} \right)$
Substituting the value of A and B from what we have assume in starting, we get
${{\tan }^{-1}}x+{{\tan }^{-1}}y=ta{{n}^{-1}}\left( \dfrac{x+y}{1-xy} \right)$
 This is the required answer.

Note: One more approach to solve this is by contradiction. That is we know the formula ${{\tan }^{-1}}x+{{\tan }^{-1}}y=ta{{n}^{-1}}\left( \dfrac{x+y}{1-xy} \right)$, considering this and proving that here xy < 1.