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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 ??

Arun
25763 Points
one year ago
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
Samyak Jain
333 Points
one year ago
1 – cosx = 2 sin2 (x/2)  and  1 + cosx = 2 cos2 (x/2)
So, (1 – cosx) / (1 + cosx) = tan2 (x/2)
Expression becomes log$\dpi{80} \sqrt{tan^2 (x/2))}$ = 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.
Raj
17 Points
one year ago
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