since there's only one side that needs manipulating, I'd pick the left side. :-)
since
1+tan = (sin+cos)/cos and
1-cot = (sin-cos)/sin
we have
[(sin+cos)*cos/(sin+cos)]^2 + [(sin-cos)*sin/(sin-cos)]^2
= sin^2+cos^2
= 1
I can only manipulate one side of the equation and the end result should be equal to the other side. please help.
((sinx + cosx)/(1 + tanx))^2 + ((sinx - cosx)/(1 - cotx))^2 = 1
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