Gievn that tan x = cot(x) - 2cot(2x),
Show that 2tan20 + 4tan40 + 8tan80 = 9(cot10 - tan10)
$\displaystyle 2\tan{20} + 4\tan{40} + 8\tan{80}$
$\displaystyle = 2(\cot{20} - 2\cot{40}) + 4(\cot{40} - 2\cot{80}) + 8(\cot{80} - 2\cot{160})$
$\displaystyle = 2\cot{20} - 4\cot{40} + 4\cot{40} - 8\cot{80} + 8\cot{80} - 16\cot{160}$
$\displaystyle = 2\cot{20} - 16\cot{160}$
$\displaystyle = 2\cot{20} - \frac{16}{\tan{160}}$.
Since $\displaystyle 160^\circ$ is in the second quadrant, $\displaystyle \tan{160} = -\tan{20}$.
Continuing...
$\displaystyle = 2\cot{20} + \frac{16}{\tan{20}}$
$\displaystyle = 2\cot{20} + 16\cot{20}$
$\displaystyle = 18\cot{20}$.
Now we use the double angle identity for cotangent.
$\displaystyle \cot{2\theta} = \frac{\cot^2{\theta} - 1}{2\cot{\theta}}$.
So $\displaystyle \cot{20} = \cot{(2\cdot 10)} = \frac{\cot^2{10} - 1}{2\cot{10}}$.
Continuing...
$\displaystyle = 18\left(\frac{\cot^2{10} - 1}{2\cot{10}}\right)$
$\displaystyle = 18\left(\frac{1}{2}\cot{10} - \frac{1}{2}\tan{10}\right)$
$\displaystyle = 9(\cot{10} - \tan{10})$.
why do i have to change $\displaystyle [2\cot{20} - \frac{16}{\tan{160}}]$ to $\displaystyle
[2\cot{20} + \frac{16}{\tan{20}}]$, is it possible work with the tan120 below?
if it's not possible, then may i would like to ask if it is true that i have to change all the angles to acute?