Question

If $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi$, then the value of $x + y + z - xyz$ is

• A. $1$
• B. $0$
• C. $-1$
• D. $\frac{1}{2}$

Hint:

$\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi$ implies $x + y + z = xyz$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Use the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) + \pi$ if $xy>1$.
Step 2: Detailed Explanation:
We know $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi$. Let $A = \tan^{-1}x$, $B = \tan^{-1}y$, $C = \tan^{-1}z$. Then $\tan(A+B+C) = \tan\pi = 0$.
$\tan(A+B+C) = \frac{\tan(A+B) + \tan C}{1 - \tan(A+B)\tan C} = \frac{\frac{x+y}{1-xy} + z}{1 - \frac{x+y}{1-xy}z} = \frac{x+y+z - xyz}{1 - xy - yz - zx} = 0$.
Thus numerator $x + y + z - xyz = 0$.
Step 3: Final Answer:
$x + y + z - xyz = 0$.