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

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Answer 1 · 2016-05-02T22:30:19

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

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

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

$=cos(theta)*1$

$=cos(theta)$

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