Question

If \( \tan^{-1}x + \tan^{-1}y = \frac{\pi}{4} \), then what is the value of \(x + y + xy\)?

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(-1\)

Hint:

For problems involving \(\tan^{-1}x + \tan^{-1}y\), convert them using \[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\frac{x+y}{1-xy}\right). \] This often simplifies the expression immediately.

Answer 1

The Correct Option is B

Concept: For inverse tangent addition, \[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\!\left(\frac{a+b}{1-ab}\right) \] provided the angles lie in the principal range.

Step 1: Apply the identity. \[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\frac{x+y}{1-xy}\right) \] Given \[ \tan^{-1}x + \tan^{-1}y = \frac{\pi}{4} \] so \[ \tan^{-1}\!\left(\frac{x+y}{1-xy}\right) = \frac{\pi}{4} \]

Step 2: Take tangent on both sides. \[ \frac{x+y}{1-xy} = \tan\left(\frac{\pi}{4}\right) = 1 \]

Step 3: Solve the equation. \[ x+y = 1-xy \] \[ x+y+xy = 1 \] \[ \boxed{1} \]