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Match the right and left sides (sinx/1-cosx) + (sinx/1+cosx) = 2cscx
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@RadEn
(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
I got (sinx(1+cosx) + sinx(1-cosx))/((1-cosx)(1+cosx))= (sinx+sinxcosx+sinx-sinxcosx)/1-cos^2x= 2sinx/1-cosx
I just don't know what to do after 2sinx/1-cos^2x
@SolomonZelman
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you mean, in the last step is 2sinx/1-cos^2 x ?
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
Yeah that's what I meant! Thank you!! :)
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