Well...I'm done so ill post as well
Anyway this would have to be broken into steps because it's quite long compared to the ones we get now....seems like an old paper
Let's deal with the numerator first
sin
2x -cos
2x
sin
2x can be written as cos
2x X tan
2x
that is because tanx=sinx/cosx
now the numerator looks like this
cos
2x X tan
2x - cos
2x
taking cos
2x common you get
cos
2x ( tan
2x - 1)
Now leave it at this. Move to the denominator
1+ 2sinxcosx
we know that sinx=cosxtanx
let's replace sinx we get
1+ 2 cosxtanx X cosx
1+ 2cos
2xtanx
we also know
sin
2x + cos
2x =1
let's replace 1 with this
sin
2x + cos
2x + cos
2xtanx
Now we still need to replace sin
2x again by cos
2x tan
2x
Doing so you get
cos
2x tan
2x+ cos
2x + 2cos
2xtanx
Now take cos
2x common
it reduces to
cos
2x (1+ tan
2x + 2tanx)
Now bring in the numerator again
cos
2x ( tan
2x - 1)
divide the two
cos
2x ( tan
2x -1)
----------------------------------------
cos
2x (1+ tan
2x + 2tanx)
cos
2x gets cancelled
remaining bring it down again
( tan
2x - 1)-------> (tanx+1)(tanx-1) 1)
------------------------
(1+ tan
2x + 2tanx)------>(tanx+1)
2 2)
Divide 1 by 2
(tanx+1)(tanx-1)
-----------------
(tanx+1) (tanx+1)
(tanx+1)(tanx-1)
------------------------
(tanx+1)
(tanx+1)You have your answer
(tanx-1)
--------
(tanx+1)