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

Question

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

Dr. Abubacker · Accepted Answer

(b)
We have$,	an{(\alpha + \beta) = \frac{	an{\alpha + 	an\beta}}{1 - 	an{\alpha	an\beta}}}$
$\because	an{\alpha = \frac{1}{1 + 2^{- x}}and	an{\beta = \frac{1}{1 + 2^{x + 1}}}}$
$	herefore	an{(\alpha + \beta) = \frac{\frac{1}{1 + \frac{1}{2^{x}}} + \frac{1}{1 + 2^{x + 1}}}{1 - \frac{1}{1 + \frac{1}{2^{x}}} \bullet \frac{1}{1 + 2^{x + 1}}}}$
$\Rightarrow 	an{(\alpha + \beta) = \frac{2^{x} + {2 \bullet 2}^{2x} + 2^{x} + 1}{1 + 2^{x} + {2 \bullet 2}^{x} + {2 \bullet 2}^{2x} - 2^{x}}}$
$\Rightarrow 	an{(\alpha + \beta) = 1}$
$\Rightarrow \ \alpha + \beta = \frac{\pi}{4}$

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

Explanation

Reviewed by Dr. Abubacker

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