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)}$Multiply numerator and denominator by $(2^x + 1)(1 + 2^{x+1})$:$	an(\alpha + \beta) = \frac{2^x(1 + 2^{x+1}) + (2^x + 1)}{(2^x + 1)(1 + 2^{x+1}) - 2^x}$$= \frac{2^x + 2^x \cdot 2^{x+1} + 2^x + 1}{2^x + 2^x \cdot 2^{x+1} + 1 + 2^{x+1} - 2^x}$$= \frac{2 \cdot 2^x + 2 \cdot 2^{2x} + 1}{2 \cdot 2^{2x} + 2^{x+1} + 1}$Since $2^{x+1} = 2 \cdot 2^x$,the numerator is $2 \cdot 2^x + 2 \cdot 2^{2x} + 1$ and the denominator is $2 \cdot 2^{2x} + 2 \cdot 2^x + 1$.Thus,$	an(\alpha + \beta) = 1$.Therefore,$\alpha + \beta = \frac{\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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