# SOLUTION: what is the derivative of -CSCxCotx? I know how to start it but i dont know how to symplify the last step CSCx((CSCx)^2+(Cotx)^2)?

Algebra ->  Algebra  -> Trigonometry-basics -> SOLUTION: what is the derivative of -CSCxCotx? I know how to start it but i dont know how to symplify the last step CSCx((CSCx)^2+(Cotx)^2)?      Log On

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 Question 163943: what is the derivative of -CSCxCotx? I know how to start it but i dont know how to symplify the last step CSCx((CSCx)^2+(Cotx)^2)?Answer by Edwin McCravy(8908)   (Show Source): You can put this solution on YOUR website!what is the derivative of -CSCxCotx? I know how to start it but i dont know how to simplify the last step CSCx((CSCx)^2+(Cotx)^2)? ``` You have demonstrated the fact that there are sometimes many ways to "play around" with the trigonometric identities, and one trigonometric expression may not necessarily be better or worse, nor simpler or more complicated, than another. Oftentimes it is just a matter of personal preference. I'll start from scratch even though your derivative is perfectly correct: Leave the negative sign factored out: Now remove the parentheses: Now factor out a CSCx That's where you got to. That may very well be simple enough as it is. Some people might argue we should write the in terms of so that it would involve only one trig function, the cosecant, like this: Then someone else may argue that the secants, cosecant, and cotangents are not as desirable as sines and cosines. They would perhaps change everything to sines and cosines: Some may consider that simpler. Others would say the form you got to was simpler because there were no fractions. Still others might say the one in sines and cosines would be simpler if it were all in just sines with no cosines, so they would replace the by . There is no hard fast rule in all cases as to what trigonometric form is considered the best or the simplest. Edwin```