#22 Measure Your Calibre

$\tan(2y) = \dfrac{2\tan y}{1 -\tan^2y} \\ \boxed{y = 2x} \\ \tan(4x) = \dfrac{2\tan(2x)}{1-\tan^2(2x)} \\ = \dfrac{2 \times \left( \dfrac{2\tan x}{1 -\tan^2x} \right)}{ 1 - \left( \dfrac{2\tan x}{1 -\tan^2x} \right)^2} \\ = \dfrac{4\tan x}{1 - \tan^2x} \times \dfrac{1}{\dfrac{(1 - \tan^2x)^2 - 4\tan^2x}{(1 - \tan^2x)^2}} \\ = \dfrac{4\tan x}{1 - \tan^2x} \times \dfrac{(1 - \tan^2x)^2}{(1 - \tan^2x)^2 - 4\tan^2x} \\ = \dfrac{4\tan x (1 - \tan^2x)}{(1 - \tan^2x)^2 - 4\tan^2x} ~~ \boxed{\text{cancel out }(1 - \tan^2x)^2} \\ = \dfrac{4\tan x(1-\tan^2x)}{1 - 2tan^2x + \tan^4x - 4\tan^2x} \\ = \dfrac{4\tan x(1-\tan^2x)}{1 - 6tan^2x + \tan^4x} \\ = \dfrac{4\tan x(1-\tan^2x)}{1 + 1\tan^4x - 6\tan^2x} \\ = \dfrac{\color{#D61F06}{4} \color{#333333}{} \tan x(1-\tan^2x)}{\color{#3D99F6}{1} \color{#333333}{} + \color{#E81990}{1} \color{#333333}{} \tan^4x - \color{#20A900}{6} \color{#333333}{} \tan^2x} \\ \boxed{\therefore \color{#D61F06}{a} + \color{#3D99F6}{b} + \color{#E81990}{k} + \color{#20A900}{m} \color{#333333}{= } \color{#D61F06}{4} + \color{#3D99F6}{1} + \color{#E81990}{1} - \color{#20A900}{6} \color{#333333}{}= 6 - 6 = 0}$

$\begin{aligned} \tan(4x) &= \dfrac{2 \tan(2x)}{1 - \tan^2(2x)} \\ &= \dfrac{2 \left( \frac{2 \tan x}{1- \tan^2 x} \right)}{1 - {\left( \frac{2 \tan x}{1- \tan^2 x} \right)}^2 } \\ &= \dfrac{ \frac{4 \tan x}{1- \tan^2 x} }{ 1 - \frac{4 \tan^2 x}{ {(1-\tan^2 x)}^2 } } \\ &= \dfrac{4 \tan x (1 - \tan^2 x)}{ {(1-\tan^2 x)}^2 - 4 \tan^2 x } \\ &= \dfrac{4 \tan x (1 - \tan^2 x)}{1 + \tan^4 x - 6 \tan^2 x} \end{aligned} \\ \implies a+b+k+m = 4+1+(-6)+1 = \boxed{0}$