# What is the derivative of csc (t/2)?

Using the quotient rule and trigonometric identities, we can obtain the
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Q&A Related to "What is the derivative of csc (t/2)"
 csc (π/2) = 1/[sin (π/2) = 1/1 = 1. that is really fundamental.you should know this ! ¶. http://answers.yahoo.com/question/index?qid=201201...
 I assume the expression is cot^2 x / ( csc^2 x - csc x) express it in terms of sin x and cos x: =cos^2 x / sin^2 x) / (1/sin^2 x - 1/sin x) =cos^2 x / sin^2 x) / [1 - sin x)/sin^2 http://wiki.answers.com/Q/How_do_you_simplify_this...
 The expression `(csc^2 A-1)/(cot A*csc A)` has to be simplified. `(csc^2 A-1)/(cot A*csc A)` => `((1/(sin^2A))-1)/((cos A/sin A)*(1/sin A))` => `((1/(sin^2A))-1)/(cos A/sin^ http://www.enotes.com/homework-help/what-simplest-...
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