Question

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

Answer 1

$$\begin{gathered} \tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } \\ =\frac{{\displaystyle \frac{1}{1+2^{-x}}}+{\displaystyle \frac{1}{1+2^{x+1}}}}{1-{\displaystyle \frac{1}{(1+2^{-x})(1+2^{x+1})}}} \\ =\frac{1+2^{x+1}+1+2^{-x}}{1+2^{x+1}+2^{-x}+2^{-x+x+1}-1} \\ =\frac{2+2^{x+1}+2^{-x}}{2+2^{x+1}+2^{-x}} \\ =1 \\ \mathrm{Therefore},\alpha +\beta ={\tan }^{-1}\left(1\right)=\frac{\mathrm{\pi }}{4}. \end{gathered}$$