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

We are given the equation $\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \frac{\pi}{4}$.
Applying the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, we get:
$$\tan^{-1}\left(\frac{(1-\alpha) + (1-\beta)}{1 - (1-\alpha)(1-\beta)}\right) = \frac{\pi}{4}$$
Taking the tangent of both sides:
$$\frac{2 - \alpha - \beta}{1 - (1 - \alpha - \beta + \alpha\beta)} = \tan\left(\frac{\pi}{4}\right) = 1$$
Simplifying the denominator:
$$\frac{2 - (\alpha + \beta)}{\alpha + \beta - \alpha\beta} = 1$$
Cross-multiplying, we obtain:
$$2 - (\alpha + \beta) = \alpha + \beta - \alpha\beta$$
$$2(\alpha + \beta) - \alpha\beta = 2$$
We are also given that $\beta = \frac{1}{3\alpha}$, which implies $\alpha\beta = \frac{1}{3}$. Substituting this value into our derived equation:
$$2(\alpha + \beta) - \frac{1}{3} = 2$$
$$2(\alpha + \beta) = 2 + \frac{1}{3} = \frac{7}{3}$$
To find the value of $6(\alpha + \beta)$, we multiply both sides of $2(\alpha + \beta) = \frac{7}{3}$ by 3:
$$3 \times [2(\alpha + \beta)] = 3 \times \frac{7}{3}$$
$$6(\alpha + \beta) = 7$$