Question

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

Answer 1

Please see below.

Explanation:

For simplicity we take,

`cotx-tanx=cosx/sinx-sinx/cosx`

`color(white)(cotx-tanx)=(cos^2x-sin^2x)/(sinxcosx)`

`color(white)(cotx-tanx)=(cos2x)/(2sinxcosx)xx2`

`color(white)(cotx-tanx)=(2cos2x)/(sin2x)`

`color(white)(cotx-tanx)=2cot2x`

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

Put `x=phi,2phi,4phi` , into `(A)` 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))`

`=tanphi+2tan2phi+2(color(violet)(cot2phi-tan2phi))`

`=tanphi+cancel(2tan2phi)+2cot2phi-cancel(2tan2phi)`

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

`=tanphi+color(blue)(cotphi-tanphi)`

`=cotphi`

`=RHS`

Answer 2

Another approach.

Explanation:

`cot2x=1/(tan2x)=(1-tan^2x)/(2tanx)`

`LHS=tanx+2tan2x+4tan4x+8cot8x`

`=tanx+2[tan2x+2tan4x+4cot8x]`

`=tanx+2[tan2x+2tan4x+4xx(1-tan^2(4x))/(2tan4x)]`

`=tanx+2[tan2x+(cancel(2tan^2(4x))+2cancel(-2tan^2(4x)))/(tan4x)]`

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

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

`=tanx+cancel(2)*(1-tan^2x)/(cancel(2)tanx)`

`=(tan^2x+1-tan^2x)/tanx=cotx=RHS`