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Mathematics 19 Online
OpenStudy (anonymous):

Prove: (sinx-cosx)/(sinx+cosx)= (2sin^2x-1)/(1+2sinxcosx)

OpenStudy (anonymous):

\[Prove: (sinx-cosx)/(sinx+cosx)= (2\sin^2x-1)/(1+2sinxcosx)\]

OpenStudy (anonymous):

Prove: (2tanx)/(1-tan^2x)+1/(2cos^2x-1) = (sinx+cosx)/(sinx-cosx) :)

OpenStudy (anonymous):

multiply left side by (sinx+cosx)/(sinx+cosx) and you get (sin^2x -cox^2x)/(1+2sinxcosx)=(sin^2x-1+sin^2x)/((1+2sinxcosx)= =(2sin^2x-1)/(1+2sinxcosx)

OpenStudy (anonymous):

ohhhh thank you :DD

OpenStudy (anonymous):

how about the second one ?

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