#### The result  $\tan^{-1}x-\tan^{-1}\left ( \frac{x-y}{1+xy} \right )$ is true when value of xy is _________.

The result $\tan^{-1}x-\tan^{-1}\left ( \frac{x-y}{1+xy} \right )$ is true when value of xy is > -1.

We have,

$\tan^{-1}x-\tan^{-1}=\tan^{-1} \frac{x-y}{1+xy}$

Principal range of tan-1a is  $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$

Let tan-1x = A and tan-1y = B … (1)

So, A,B  $\epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$

We know that, $\tan\left ( A-B \right )=\frac{\tan A - \tan B}{1-\tan A \tan B }$  … (2)

From (1) and (2), we get,

Applying, tan-1 both sides, we get,

$\tan^{-1}\tan\left ( A-B \right )=\tan^{-1}\frac{x-y}{1-xy}$

As, principal range of tan-1a is $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$

So, for tan-1tan(A-B) to be equal to A-B,

A-B must lie in  $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$– (3)

Now, if both A,B < 0, then A, B  $\epsilon \left ( -\frac{\pi}{2},0\right )$

∴ A $\epsilon \left ( -\frac{\pi}{2},0\right )$   and -B $\epsilon \left ( 0,\frac{\pi}{2}\right )$

So, A – B  $\epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$

So, from (3),

tan-1tan(A-B) = A-B

$\Rightarrow \tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{z+xy}$

Now, if both A,B > 0, then A, B $\epsilon \left ( 0,\frac{\pi}{2}\right )$

∴ A $\epsilon \left ( 0,\frac{\pi}{2}\right )$  and -B  $\epsilon \left ( -\frac{\pi}{2},0\right )$

So, A – B  $\epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$

So, from (3),

tan-1tan(A-B) = A-B

$\Rightarrow \tan{-1}x-\tan{-1}y=\tan^{-1}\frac{x-y}{z+xy}$

Now, if A > 0 and B < 0,

Then, A $\epsilon \left ( 0,\frac{\pi}{2}\right )$   and B  $\epsilon \left ( 0,\frac{\pi}{2}\right )$

∴ A  $\epsilon \left ( 0,\frac{\pi}{2}\right )$ and -B  $\epsilon \left ( 0,\frac{\pi}{2}\right )$

So, A – B $\epsilon$ (0,π)

But, required condition is A – B $\epsilon$ $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$

As, here A – B $\epsilon$ (0,π), so we must have A – B $\epsilon$ $\left ( 0,\frac{\pi}{2} \right )$

$A-B< \frac{\pi}{2}$

$A< \frac{\pi}{2} +B$
Applying tan on both sides,

$\tan A< \tan\left ( \frac{\pi}{2} +B \right )$

As, $\tan\left ( \frac{\pi}{2} +\alpha \right )=-\cot \alpha$

So, tan A < - cot B

Again,  $\cot \alpha=\frac{1}{\tan \alpha}$

So, $\tan A< \frac{1}{\tan B}$

⇒ tan A tan B < -1

As, tan B < 0

xy > -1

Now, if A < 0 and B > 0,

Then, A $\epsilon$ $\left ( -\frac{\pi}{2} ,0\right )$  and B $\epsilon$ $\left ( 0,\frac{\pi}{2} \right )$

∴ A $\epsilon$   $\left ( -\frac{\pi}{2} ,0\right )$ and -B $\epsilon$ $\left ( -\frac{\pi}{2} ,0\right )$

So, A – B $\epsilon$ (-π,0)

But, required condition is A – B $\epsilon$ $\left ( -\frac{\pi}{2} ,\frac{\pi}{2}\right )$

As, here A – B $\epsilon$ (0,π), so we must have A – B $\epsilon$ $\left ( -\frac{\pi}{2} ,0\right )$

$\Rightarrow A-B> -\frac{\pi}{2}$

$\Rightarrow A>B -\frac{\pi}{2}$

Applying tan on both sides,

$\tan A>\tan\left (B -\frac{\pi}{2} \right )$

As, $\tan\left (\alpha -\frac{\pi}{2} \right )=-\cot \alpha$

So, tan B > - cot A

Again, $\cot \alpha\frac{1}{\tan \alpha}$

So, $\tan B >-\frac{1}{\tan A}$

⇒ tan A tan B > -1

⇒xy > -1