If tan alpha = (1 + 2^{-x})^{-1} and tan beta | vedclass

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(B) Given $	an \alpha = \frac{1}{1 + 2^{-x}} = \frac{2^x}{2^x + 1}$ and $	an \beta = \frac{1}{1 + 2^{x+1}}$.Using the formula $	an(\alpha + \beta)

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$\pi / 4$

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(B) Given $	an \alpha = \frac{1}{1 + 2^{-x}} = \frac{2^x}{2^x + 1}$ and $	an \beta = \frac{1}{1 + 2^{x+1}}$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$:$	an(\alpha + \beta) = \frac{\frac{2^x}{2^x + 1} + \frac{1}{1 + 2^{x+1}}}{1 - \left(\frac{2^x}{2^x + 1}ight) \left(\frac{1}{1 + 2^{x+1}}ight)}$Simplify the numerator: $\frac{2^x(1 + 2^{x+1}) + (2^x + 1)}{(2^x + 1)(1 + 2^{x+1})} = \frac{2^x + 2^{2x+1} + 2^x + 1}{(2^x + 1)(1 + 2^{x+1})} = \frac{2 \cdot 2^x + 2 \cdot 2^{2x} + 1}{(2^x + 1)(1 + 2^{x+1})}$.Simplify the denominator: $\frac{(2^x + 1)(1 + 2^{x+1}) - 2^x}{(2^x + 1)(1 + 2^{x+1})} = \frac{2^x + 2^{2x+1} + 1 + 2^{x+1} - 2^x}{(2^x + 1)(1 + 2^{x+1})} = \frac{2^{2x+1} + 2^{x+1} + 1}{(2^x + 1)(1 + 2^{x+1})}$.Since the numerator and denominator are equal, $	an(\alpha + \beta) = 1$.Therefore, $\alpha + \beta = \pi / 4$.

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

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