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Mathematics 15 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

@RadEn

OpenStudy (solomonzelman):

(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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@SolomonZelman

OpenStudy (raden):

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

OpenStudy (raden):

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

OpenStudy (anonymous):

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

OpenStudy (raden):

welcome

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