# (1/cosec2x-1/sin2x)sin2xcos2x =1+sin2xcos2x/2-sin2xcos2x

hala718 | Certified Educator

(1/cosec2x -1/sin2x)sin2xcos2x = 1+sin2xcos2x/2-sin2xcos2x

Let us start from the left side:

We know that cosec= 1/sin

==> (sin2x -1/sin2x)sin2xcos2x

Open brackets:

==> [(sin2x)^2 cos2x]-1/cos2x

==> (sin^2(2x)cos^2(2x)-1)/cos2x)

=(sin(2x)cos(2x)-1)(sin(2x)cos(2x)+1)/ (cos^2(x)-sin^2 (x))

neela | Student

To prove that

(1/cosec2x-1/sin2x)sin2xcos2x = (1+sin2xcos2x)/(2-sin2xcos2x)

Since  1/cosec2x = sin2x, we have to prove that

(sin2x-1/sin2x)sin2xcos2x = (1+sin2xcos2x)(2-sin2xcos2x)

Solution:

since 1/cosex = sin2x , the LHS could be simplified to

{(sin2x)^2 - 1}cos2x  = -{1 - (sin2x)^2}cos2x = -(cos2x)^3

For x = 0. LHS value = -(cos0)^3 = -1

RHS: For x= 0

= {1+sin2xcos2x)/(2-sin2xcos2x) = {1+0*1}/{2-0*1} =1/2.

So is not an identity.

giorgiana1976 | Student

First, the rule of brackets in mathematics have to be followed!

[(1/cosec2x)-(1/sin2x)]*sin2xcos2x=(1+sin2xcos2x)/(2-sin2xcos2x), then, we'll do the following:

1/cosec2x = (sin 2x)^2

[(1/cosec2x)-(1/sin2x)]=[(sin 2x)^2-(1/sin2x)]

[(sin 2x)^2-(1/sin2x)] = [(sin 2x)^3 - 1]/sin 2x

[(1/cosec2x)-(1/sin2x)]*sin2xcos2x=[(sin 2x)^3 - 1]*sin2xcos2x/sin 2x

After reducing similar terms, we'll have:

[(sin 2x)^3 - 1]*cos 2x

To the right side, we'll have:(1+sin2xcos2x)/(2-sin2xcos2x)

(1+sin2xcos2x)/(2-sin2xcos2x)={1+2sinx*cosx[1-2(sinx)^2]}/{2-2sinx*cosx[1-2(sinx)^2}