1. ## Substitution? Confused!

Gievn that tan x = cot(x) - 2cot(2x),

Show that 2tan20 + 4tan40 + 8tan80 = 9(cot10 - tan10)

2. Originally Posted by Punch
Gievn that tan x = cot(x) - 2cot(2x),

Show that 2tan20 + 4tan40 + 8tan80 = 9(cot10 - tan10)
$2\tan{20} + 4\tan{40} + 8\tan{80}$

$= 2(\cot{20} - 2\cot{40}) + 4(\cot{40} - 2\cot{80}) + 8(\cot{80} - 2\cot{160})$

$= 2\cot{20} - 4\cot{40} + 4\cot{40} - 8\cot{80} + 8\cot{80} - 16\cot{160}$

$= 2\cot{20} - 16\cot{160}$

$= 2\cot{20} - \frac{16}{\tan{160}}$.

Since $160^\circ$ is in the second quadrant, $\tan{160} = -\tan{20}$.

Continuing...

$= 2\cot{20} + \frac{16}{\tan{20}}$

$= 2\cot{20} + 16\cot{20}$

$= 18\cot{20}$.

Now we use the double angle identity for cotangent.

$\cot{2\theta} = \frac{\cot^2{\theta} - 1}{2\cot{\theta}}$.

So $\cot{20} = \cot{(2\cdot 10)} = \frac{\cot^2{10} - 1}{2\cot{10}}$.

Continuing...

$= 18\left(\frac{\cot^2{10} - 1}{2\cot{10}}\right)$

$= 18\left(\frac{1}{2}\cot{10} - \frac{1}{2}\tan{10}\right)$

$= 9(\cot{10} - \tan{10})$.

3. why do i have to change $[2\cot{20} - \frac{16}{\tan{160}}]$ to $
[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?

4. Originally Posted by Punch
why do i have to change $[2\cot{20} - \frac{16}{\tan{160}}]$ to $
[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?
Yes you have to change all the angles to acute.

It would make sense, since you're trying to get all the angles to $10^\circ$...

5. Hello everyone

A slight variation at the end:

Once you get to $18\cot20$, use the original result, re-arranged as:
$2\cot2x= \cot x -\tan x$

$\Rightarrow 18\cot20=9(\cot10 -\tan10)$