Match the right and left sides
(sinx/1-cosx) + (sinx/1+cosx) = 2cscx

Question

Match the right and left sides
(sinx/1-cosx) + (sinx/1+cosx) = 2cscx

anonymous · Answer

@RadEn

solomonzelman · Answer

(sinx/1-cosx) + (sinx/1+cosx) = 2cscx
is same as
(sinx-cosx) + (sinx+cosx) = 2cscx
sinx-cosx + sinx+cosx = 2cscx
2sinx=2cscx
sinx=cscx
sinx=1/sinx

sinx  has to be =1
so,    x=90

anonymous · Answer

I got 
(sinx(1+cosx) + sinx(1-cosx))/((1-cosx)(1+cosx))=
(sinx+sinxcosx+sinx-sinxcosx)/1-cos^2x=
2sinx/1-cosx

anonymous · Answer

I just don't know what to do after 2sinx/1-cos^2x

anonymous · Answer

@SolomonZelman

raden · Answer

you mean, in the last step is 2sinx/1-cos^2 x ?

raden · Answer

if yes, you are right.
just simplify again. use the identity :
1 - cos^2 x = sin^2 x


therefore, it can be
2sinx/1-cos^2x = 2sinx/sin^2 x = 2/sinx = 2cscx

anonymous · Answer

Yeah that's what I meant! Thank you!! :)

raden · Answer

welcome