MATHS

Asked by NawaarStart off by using tan^2(A/2) = ( sin^2(A)/(1+cosA)^2) So tan^2(x/2) = ( sin^2(x)/(1+cosx)^2) = ( 1- cos^2(x))/(1+cosx)^2 = (1-cosx)(1+cosx)/((1+cosx)^2) Cancelling term gives (1-cosx)/(1+cosx) = (secx-1)/(1+secx) LHS=RHS

When proving such relations we have a choice, that is we can either start from the left hand side (LHS) and show that it equals the right hand side (RHS) or vice versa. I will start with the RHS, RHS = tan^2(x/2), = {sin^2(x/2)} / {cos^2(x/2)}, = {(1/2)(1+cos(x))} / {(1/2)(1-cos(x))} = {1+cos(x)} / {1-cos(x)}, = {1+(1/sec(x))} / {1-(1/sec(x))}, = (sec(x)+1) / (sec(x)-1), = LHS. Write this down on a piece of paper, and try to verify each step for yourself.

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