First slide

Compound angles

Question

$If \tan α = {(1 + 2^{- x})}^{- 1} and \tan β = {(1 + 2^{x + 1})}^{- 1} . Then α + β =$

Difficult

A

$\frac{π}{6}$
B

$\frac{π}{4}$
C

$\frac{π}{3}$
D

$\frac{π}{2}$

Solution

$\begin{array}{l} W e h a v e \tan α = {(\frac{2^{x} + 1}{2^{x}})}^{- 1} \\ \Rightarrow \tan α = \frac{2^{x}}{2^{x} + 1} \\ A l s o \tan β = {(1 + 2^{x} \cdot 2)}^{- 1} \\ \Rightarrow \tan β = \frac{1}{1 + 2^{x} \cdot 2} \\ N o w \tan (α + β) = \frac{\frac{2^{x}}{2^{x} + 1} + \frac{1}{1 + 2^{x} \cdot 2}}{1 - \frac{2^{x}}{2^{x} + 1} \cdot \frac{1}{1 + 2^{x} \cdot 2}} \end{array}$
$\begin{matrix} \Rightarrow \tan (α + β) = \frac{\frac{t}{t + 1} + \frac{1}{2 t + 1}}{1 - (\frac{t}{t + 1}) (\frac{1}{1 + 2 t})} (w h e r e t = 2^{x}) \\ = \frac{(2 t + 1) t + t + 1}{(t + 1) (1 + 2 t) - t} \\ = \frac{2 t^{2} + t + t + 1}{t + 2 t^{2} + 1 + 2 t - t} = 1 \\ \Rightarrow α + β = \frac{π}{4} \end{matrix}$

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