Question

prove that sec 8x -1\sec4x-1=tan8x\tan2x

Answer 1

We need to prove the following :-

[sec(8x) - 1] / [sec(4x) - 1]  =  tan(8x) / tan(2x) --------- I

So now, if we prove that

[sec(8x) - 1] * [tan(2x)]  =  [sec(4x) - 1] * tan(8x) ------- II

Then by cross multiplication law, we have also proved I.

[1 - cos(8x)] / cos(8x) * sin(2x) / cos(2x)  =  [1 - cos(4x)] / cos(4x) * sin(8x) / cos(8x)   -------- III

We substituted sec(8x) as 1 / cos(8x) in II to get III

Now, III reduces to

[1 - cos(8x)]  * sin(2x) / cos(2x)  =  [1 - cos(4x)] / cos(4x) * sin(8x)

since cos(8x) is canceled.

Now, we multiply the terms.

tan(2x) - [cos(8x) * tan(2x)]  =  [sin(8x) / cos(4x)] - sin(8x)

Now we use this formula for sin(2A) = 2 [sinA * cosA]

Here A = 8x

tan(2x) - [cos(8x) * tan(2x)]  =  2[sin(4x) * cos(4x) / cos(4x)] - sin(8x)

This simplifies to

tan(2x) - [cos(8x) * tan(2x)]  =  [2 sin(4x)] - sin(8x)

tan(2x) { 1 - cos(8x) } = 2 sin(4x) - 2 sin(4x) * cos(4x)

tan(2x) { 1 - cos(8x) } = 2 sin(4x) { 1 - cos(4x) }

tan(2x) { 1 - cos(8x) } = 4 sin(2x) * cos(2x) { 1 - cos(4x) }

Substitute tan(2x) = sin(2x) / cos(2x)

{ 1 - cos(8x) }  =  4 $cos^{2}$(2x) { 1 - cos(4x) } ------- IV

Now using cos(2A) = 2$cos^{2}$A - 1

Substitute this in IV with A = 8x and then 4x

[2 - 2 $cos^{2}$(4x)] / [2 - 2

$sin^{2}$(4x) /

$[2 sin(2x) * cos(2x)]^{2}$ / $sin^{2}$(2x)

4 $cos^{2}$(2x) ------ LHS

Now RHS is also 4 $cos^{2}$(2x).

So the condition is for I to be true, here LHS must be equal to RHS.

Since this is satisfied.

[sec(8x) - 1] / [sec(4x) - 1]  =  tan(8x) / tan(2x) is true

Hence Proved.