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

2 Answers
Jun 5, 2018

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

Jun 5, 2018

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