Question

If $\tan\alpha = (1 + 2^{- x})^{- 1},$ $\tan\beta = (1 + 2^{x + 1})^{- 1}$, then $\alpha + \beta$ equals

• A. $\pi/6$
• B. $\pi/4$
• C. $\pi/3$
• D. $\pi/2$

Answer 1

Answer: (B)

$\pi/4$

Answer 2

$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

⇒ $\tan(\alpha + \beta) = \frac{\frac{1}{1 + \frac{1}{2^{x}}} + \frac{1}{1 + 2^{x + 1}}}{1 - \frac{1}{1 + 1/2^{x}}\frac{1}{1 + 2^{x + 1}}}$

⇒ $\tan(\alpha + \beta) = \frac{2^{x} + 2.2^{x + x} + 2^{x} + 1}{1 + 2^{x} + 2.2^{x} + 2.2^{x + x} - 2^{x}}$

⇒ $\tan(\alpha + \beta) = 1 \Rightarrow \alpha + \beta = \frac{\pi}{4}$.