Question

Let $0<\alpha<1, \beta = \frac{1}{3\alpha}$ and $\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \frac{\pi}{4}$. Then $6(\alpha + \beta)$ is equal to:

• A. 6
• B. 7
• C. 8
• D. 9

Hint:

Use the identity $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ and substitute the relationship $\alpha\beta = 1/3$ into the resulting expression.

Answer 1

The Correct Option is B

Let $x = 1 - \alpha$ and $y = 1 - \beta$. The equation is $\tan^{-1} x + \tan^{-1} y = \frac{\pi}{4}$.
This implies that $\tan(\tan^{-1} x + \tan^{-1} y) = \tan(\frac{\pi}{4}) = 1$.
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$, we have:
$$\frac{x + y}{1 - xy} = 1 \implies x + y = 1 - xy \implies x + y + xy = 1$$
Substitute $x$ and $y$ back in terms of $\alpha$ and $\beta$:
$$(1 - \alpha) + (1 - \beta) + (1 - \alpha)(1 - \beta) = 1$$
$$2 - \alpha - \beta + 1 - \alpha - \beta + \alpha\beta = 1$$
$$3 - 2(\alpha + \beta) + \alpha\beta = 1$$
$$2(\alpha + \beta) - \alpha\beta = 2$$
Given $\beta = \frac{1}{3\alpha}$, we have $\alpha\beta = \frac{1}{3}$. Plugging this in:
$$2(\alpha + \beta) - \frac{1}{3} = 2 \implies 2(\alpha + \beta) = \frac{7}{3}$$
Multiplying the entire equation by 3 gives:
$$6(\alpha + \beta) = 7$$