Question

If \(x=\tan A, y=\tan B, z=\tan C\) and \[ xy+yz+zx=1, \] then evaluate \[ \frac{(1-x^2)(1-y^2)(1-z^2)}{(1+x^2)(1+y^2)(1+z^2)} \]

• A. \(\frac{4xy}{(1+x^2)(1+y^2)}\)
• B. \(4xyz\)
• C. \(\frac{4xy}{(1+x^2)(1+y^2)}+\frac{2z}{1+z^2}\)
• D. \(1\) \bigskip

Hint:

Whenever \(xy+yz+zx=1\) for tangents, it often implies \(A+B+C=\frac{\pi}{2}\).

Answer 1

The Correct Option is D

Concept: Use identity: \[ \frac{1-\tan^2\theta}{1+\tan^2\theta}=\cos2\theta \]

Step 1: Convert expression.
\[ \frac{1-x^2}{1+x^2}=\cos2A \] Similarly for all. \[ \Rightarrow \cos2A\cos2B\cos2C \]

Step 2: Use condition.
Given: \[ \tan A\tan B+\tan B\tan C+\tan C\tan A=1 \Rightarrow A+B+C=\frac{\pi}{2} \]

Step 3: Apply identity.
\[ \cos2A\cos2B\cos2C=1 \] \[ \boxed{(D)} \]