SOLUTION: Prove using reciprocal and pythagorean identities: (cscx-cotx)/(secx-1)= cotx I have already done: ((1/sinx)-(cosx/sinx))/((1/cosx)-1))= ((1-cosx)/sinx))/((1-cosx)/(cosx))= ((1

Algebra ->  Trigonometry-basics -> SOLUTION: Prove using reciprocal and pythagorean identities: (cscx-cotx)/(secx-1)= cotx I have already done: ((1/sinx)-(cosx/sinx))/((1/cosx)-1))= ((1-cosx)/sinx))/((1-cosx)/(cosx))= ((1      Log On


   

Question 826390: Prove using reciprocal and pythagorean identities: (cscx-cotx)/(secx-1)= cotx
I have already done:
((1/sinx)-(cosx/sinx))/((1/cosx)-1))=
((1-cosx)/sinx))/((1-cosx)/(cosx))=
((1-cosx)/sinx))*((cosx)/(1-cosx))=
((1-cosx)cosx))/((sinx(1-cosx))=
((cosx-(cos^2))/(sinx-sinxcosx)
This is where is get stuck...
Thank you
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
You're going to laugh (or cry) when you see how close you were:
%28cos%28x%29-cos%5E2%28x%29%29%2F%28sin%28x%29-sin%28x%29cos%28x%29%29
We're going to reduce this fraction. And since only factors cancel, we will factor the numerator and denominator:
%28cos%28x%29%281-cos%28x%29%29%29%2F%28sin%28x%29%281-cos%28x%29%29%29
And, as we can see, a factor will cancel:

leaving
cos%28x%29%2Fsin%28x%29
which is equal to:
cot%28x%29