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QUESTION : Differentiate with respect to x.ANSWER :COSECXHow to do this..plz help What we should do either use log formula or any other ??
Dear student First solve 1 – cos x / 1 + cos x = 2 sin^2 (x/2) / 2 cos^2 (x/2) expression becomes = log(tanx/2) now on differentiation = cosx
1 – cosx = 2 sin2 (x/2) and 1 + cosx = 2 cos2 (x/2)So, (1 – cosx) / (1 + cosx) = tan2 (x/2)Expression becomes log = log(tan(x/2))d{log(tan(x/2))}/dx = {1/tan(x/2)} sec2(x/2) . (1/2) = 1 / 2sin(x/2) cos(x/2) = 1 / sinx = cosecx.
1 - cosx = 2sin^2 (x/2) and 1 + cos x= 2 cos^ 2(x/2)So, (1 - cosx) / (1 + cosx)= tan^ 2(x/2)Then we got log√tan^2(x/2) = log(tan(x/2))Then diff. W.R.T x We get 1/ sin x = cosec x
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