Question

Question #8e102

Answer 1

It is (*)

#sqrt((1+sintheta)/(1-sintheta))= sqrt([(1+sintheta)*(1+sintheta)]/[(1-sintheta)(1+sintheta))]=sqrt[(1+sintheta)^2/(1-sin^2theta))=sqrt[(1+sintheta)^2]/sqrt[cos^2theta]=abs(1+sintheta)/abs(costheta)=(1+sintheta)/|costheta|#

(*) Footnotes

1) In the second step we multiply both numerator and denominator with `1+sintheta`

2) Because `1>=sintheta` hence `abs(1+sintheta)=1+sintheta`