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:

Since the equation must hold for many different pairs of x and y, but the expression x + y + xy always works out to the same number, try substituting one simple valid pair first, such as y = 0, to find the value quickly, then check a second pair for confidence.

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. \[ x+y = 1-xy \] \[ x+y+xy = 1 \] \[ \boxed{1} \]

Answer 2

Concept:

• When a problem guarantees one fixed numeric value for an expression even though x and y are not individually fixed, that expression must stay the same no matter which valid pair of x, y is chosen.
• This means picking any convenient pair that satisfies the given condition, and evaluating the expression directly, is enough to find the answer.

Step 1: Choose a simple pair satisfying $\tan^{-1}x + \tan^{-1}y = \frac{\pi}{4}$.
Let $y = 0$. Then $\tan^{-1}x = \frac{\pi}{4}$, so $x = 1$.

Step 2: Evaluate the expression for this pair.
$x + y + xy = 1 + 0 + (1)(0) = 1$

Step 3: Choose a second, different pair to confirm the value stays the same.
Let $x = y$. Then $2\tan^{-1}x = \frac{\pi}{4}$, so $\tan^{-1}x = \frac{\pi}{8}$, giving $x = \tan\frac{\pi}{8} = \sqrt{2} - 1$.

Step 4: Evaluate the expression for this second pair.
$x + y + xy = 2(\sqrt{2}-1) + (\sqrt{2}-1)^2 = (2\sqrt{2} - 2) + (3 - 2\sqrt{2}) = 1$

Final Answer: $1$