Question:
If $\tan \alpha=\frac{1}{1+2^{-x}}$ and $\tan \beta=\frac{1}{1+2^{x+1}}$, then write the value of $\alpha+\beta$ lying in the interval $\left(0, \frac{\pi}{2}\right)$.
Solution:
$\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$
$=\frac{\frac{1}{1+2^{-x}}+\frac{1}{1+2^{x+1}}}{1-\frac{1}{\left(1+2^{-x}\right)\left(1+2^{x+1}\right)}}$
$=\frac{1+2^{x+1}+1+2^{-x}}{1+2^{x+1}+2^{-x}+2^{-x+x+1}-1}$
$=\frac{2+2^{x+1}+2^{-x}}{2+2^{x+1}+2^{-x}}$
$=1$
Therefore, $\alpha+\beta=\tan ^{-1}(1)=\frac{\pi}{4}$.
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