Question

prove that (sinx+cosx)/(sinx-cosx) + (sinx-cosx)/(sinx+cosx)=2/(sin2x-cos2x) ​

Answer 1

Answer:

cosx+sinx=cos2x+sin2x, then

cosx-cos2x=sin2x-sinx

2sin(x+2x)/2sin(2x-x)/2=2cos(2x+x)/2sin(2x-x)/2

sin(3x/2) sin(x/2) - cos(3x/2) sin(x/2) = 0

sin(x/2) [sin(3x/2) - cos(3x/2)] = 0

Either sin(x/2) = 0, or sin(3x/2) - cos(3x/2)=0

So if sin(x/2) = 0, then

sin (x/2) = sin 0

Thus x/2 = nπ +(-1)ⁿ ×0

or x/2 = nπ or x = 2nπ.

And if sin(3x/2) - cos(3x/2) = 0

Dividing both sides by cos(3x/2), we get

tan(3x/2) - 1 = 0

tan(3x/2) = 1

tan(3x/2) = tan(π/4)

3x/2 = nπ + π/4

3x = 2nπ + π/2

x = (2nπ/3) + π/6.

Step-by-step explanation:

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