Under trigonometry in deravatives in calculus
asked Nov 24, 2016 in Trigonometry Answers by Martin Kishao

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The question doesn't specify whether differentiation or integration is to be applied, so let's look at both.

1/(cosec(x)-cot(x))=sin(x)/(1-cos(x)), by multiplying top and bottom by sin(x).


Let y=1-cos(x) so y'=sin(x) and dy=sin(x)dx so to evaluate ∫(sin(x)/(1-cos(x))dx we can replace sin(x)dx with dy and we have ∫dy/y=ln|y|=ln|1-cos(x)| or more generally ln|A(1-cos(x)| where A is a constant.


To differentiate the expression, let u=sin(x) and v=1-cosx, so u'=cos(x) and v'=sin(x).

The derivative is (vu'-uv')/v^2=(cos(x)(1-cos(x))-sin^2(x))/(1-cos(x))^2=cos(x)-cos^2(x)-sin^2(x))/(1-cos(x))^2.

This simplifies to (cos(x)-1)/(1-cos(x))^2=-1/(1-cos(x)) or 1/(cos(x)-1).


answered Nov 25, 2016 by Rod Top Rated User (487,420 points)
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