Question

If $\tan^{-1}\left(\frac{x}{2}\right) + \tan^{-1}\left(\frac{y}{2}\right) + \tan^{-1}\left(\frac{z}{2}\right) = \frac{\pi}{2}$ then $xy + yz + zx =$}

• A. $0$
• B. $2$
• C. $-1$
• D. $4$

Hint:

If $\sum \tan^{-1} a_i = \pi/2$, then $\sum a_i a_j = 1$.

Answer 1

The Correct Option is D

Step 1: Concept
Use the identity $\tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \tan^{-1} \left( \frac{a+b+c-abc}{1-ab-bc-ca} \right)$.

Step 2: Meaning
Since the sum is $\pi/2$, the tangent of the sum is undefined, meaning the denominator must be zero.

Step 3: Analysis
Let $a = x/2, b = y/2, c = z/2$. $1 - (ab + bc + ca) = 0 \implies ab + bc + ca = 1$. $\frac{x}{2}\frac{y}{2} + \frac{y}{2}\frac{z}{2} + \frac{z}{2}\frac{x}{2} = 1$. $\frac{xy + yz + zx}{4} = 1$.

Step 4: Conclusion
$xy + yz + zx = 4$. Final Answer: (D)