Prove the following trigonometric identities by showing that the left-hand side is equivalent to the right-hand side.
a) sin(t)/1-cos(t) + 1-cos(t)/sin(t) = 2sin(t)/1-cos^2(t)
b) 2tan(t)-tan(t)(2sin^2(t))/sin(t)*cos(t) = (2-2tan^2(t)/cos^2(t)+tan^2(t)*cos^2(t)) +2sec^2(t)-2
assuming the usual carelessness with parentheses,
sint/(1-cost) + (1-cost)/sint
= (sin^2t + (1-cost)^2)/((1-cost) sint)
= (sin^2t + cos^2t - 2cost + 1)/((1-cost) sint)
= (2 - 2cost)/((1-cost) sint)
= 2/sint
= 2sint/sin^2t
= 2sint/(1 - cos^2t)
LS: 2tan(t)-tan(t)(2sin^2(t))/sin(t)*cos(t)
= 2tant(1 - 2sin^2t)/(1/2 sin2t)
= 4tantcos2t/sin2t
= 4tant cot2t
Not sure how to parse the right side. Put some parentheses in to make things clearer. Better yet, how about to make a start on it and whow whatcha got?