Question

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

Answer 1

Hint: Use the trigonometric identity related to tan that is $\tan (x+y)={\displaystyle \frac{\tan x+\tan y}{1-\tan x\tan y}}$ . 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, $\tan (x+y)={\displaystyle \frac{\tan x+\tan y}{1-\tan x\tan y}}$, so we can write,
$\tan (A+B)={\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B}}$
Now substituting the values from equation (i) and (ii), we get
$\tan (A+B)={\displaystyle \frac{x+y}{1-xy}}$
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, $\tan (A+B)={\displaystyle \frac{x+y}{1-xy}}$ is true only when xy < 1.
Now we will multiply the above expression with ${\tan }^{-1}$ , we get
$ta{n}^{-1}(\tan (A+B))=ta{n}^{-1}\left({\displaystyle \frac{x+y}{1-xy}}\right)$
We know $\tan ,{\tan }^{-1}$ gets cancelled, so we get
$A+B=ta{n}^{-1}\left({\displaystyle \frac{x+y}{1-xy}}\right)$
Substituting the value of A and B from what we have assume in starting, we get
${\tan }^{-1}x+{\tan }^{-1}y=ta{n}^{-1}\left({\displaystyle \frac{x+y}{1-xy}}\right)$
 This is the required answer.

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