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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.

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.

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