# Prove it -- tan Ø + 2 tan 2 ø + 4 tan 4 ø + 8 cot 8 ø = cot ø ?

Jun 5, 2018

#### Explanation:

For simplicity we take,

$\cot x - \tan x = \cos \frac{x}{\sin} x - \sin \frac{x}{\cos} x$

$\textcolor{w h i t e}{\cot x - \tan x} = \frac{{\cos}^{2} x - {\sin}^{2} x}{\sin x \cos x}$

$\textcolor{w h i t e}{\cot x - \tan x} = \frac{\cos 2 x}{2 \sin x \cos x} \times 2$

$\textcolor{w h i t e}{\cot x - \tan x} = \frac{2 \cos 2 x}{\sin 2 x}$

$\textcolor{w h i t e}{\cot x - \tan x} = 2 \cot 2 x$

=>color(red)( 2cot2x=cotx-tanx...to(A)

Put $x = \phi , 2 \phi , 4 \phi$ , into $\left(A\right)$ we get

color(blue)((i)2cot2phi=cotphi-tanphi

color(violet)((ii)2cot4phi=cot2phi-tan2phi

color(brown)((iii)2cot8phi=cot4phi-tan4phi

We take,

LHS=tan Ø + 2 tan 2 ø + 4 tan 4 ø + 4color(brown)((2 cot 8 ø)touse(iii))

=tan Ø + 2 tan 2 ø + 4 tan 4 ø + 4(color(brown)(cot4phi- tan4phi))

=tan Ø + 2 tan 2 ø +cancel( 4 tan 4 ø )+4cot4phi- cancel(4tan4phi)

=tan Ø + 2 tan 2 ø + 4 cot 4 ø

=tanphi+2tan2phi+2(color(violet)((2cot4phi).....................touse(ii))

$= \tan \phi + 2 \tan 2 \phi + 2 \left(\textcolor{v i o \le t}{\cot 2 \phi - \tan 2 \phi}\right)$

$= \tan \phi + \cancel{2 \tan 2 \phi} + 2 \cot 2 \phi - \cancel{2 \tan 2 \phi}$

=tanphi+color(blue)((2cot2phi).........................................touse(i)

$= \tan \phi + \textcolor{b l u e}{\cot \phi - \tan \phi}$

$= \cot \phi$

$= R H S$

Jun 5, 2018

Another approach.

#### Explanation:

$\cot 2 x = \frac{1}{\tan 2 x} = \frac{1 - {\tan}^{2} x}{2 \tan x}$

$L H S = \tan x + 2 \tan 2 x + 4 \tan 4 x + 8 \cot 8 x$

$= \tan x + 2 \left[\tan 2 x + 2 \tan 4 x + 4 \cot 8 x\right]$

$= \tan x + 2 \left[\tan 2 x + 2 \tan 4 x + 4 \times \frac{1 - {\tan}^{2} \left(4 x\right)}{2 \tan 4 x}\right]$

$= \tan x + 2 \left[\tan 2 x + \frac{\cancel{2 {\tan}^{2} \left(4 x\right)} + 2 \cancel{- 2 {\tan}^{2} \left(4 x\right)}}{\tan 4 x}\right]$

$= \tan x + 2 \left[\tan 2 x + \frac{2 \left(1 - {\tan}^{2} \left(2 x\right)\right)}{2 \tan 2 x}\right]$

=tanx+2*[tan^2(2x)+1-tan^2(2x))/(tan2x)]

$= \tan x + \cancel{2} \cdot \frac{1 - {\tan}^{2} x}{\cancel{2} \tan x}$

$= \frac{{\tan}^{2} x + 1 - {\tan}^{2} x}{\tan} x = \cot x = R H S$