If tan α = 1/(1 + 2−x) and tan β = 1/(1 + 2x), Find α + β
Question:
If
\[
\tan\alpha=\frac{1}{1+2^{-x}}
\]
and
\[
\tan\beta=\frac{1}{1+2^x}
\]
find the value of
\[
\alpha+\beta
\]
lying in the interval
\[
\left(0,\frac{\pi}{2}\right)
\]
Solution
\[ \tan\alpha = \frac{1}{1+\frac1{2^x}} = \frac{2^x}{2^x+1} \]
\[ \tan\beta = \frac{1}{2^x+1} \]
Using, \[ \tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta} {1-\tan\alpha\tan\beta} \]
\[ = \frac{ \frac{2^x}{2^x+1}+\frac1{2^x+1} } { 1-\frac{2^x}{(2^x+1)^2} } \]
\[ = \frac{1}{1} =1 \]
\[ \tan(\alpha+\beta)=1 \]
Since \[ 0<\alpha+\beta<\frac{\pi}{2} \]
\[ \boxed{ \alpha+\beta=\frac{\pi}{4} } \]