Question

How do you prove #(cos[theta]+cot[theta])/(csc[theta]+1)= cos[theta]#?

Answer 1

Using the definitions of `cot(theta)` and `csc(theta)`, for `sin(theta)!=0` and `sin(theta)!=-1`, we have

`(cos(theta)+cot(theta))/(csc(theta)+1) = (cos(theta)+cos(theta)/sin(theta))/(1/sin(theta)+1)`

`=cos(theta)(1+1/sin(theta))/(1+1/sin(theta))`

`=cos(theta)*1`

`=cos(theta)`