Question 860074
I'm only going to be altering the right side.



1+cscA=cscA(1+sinA)



1+cscA=cscA(1)+cscA(sinA) ... distribution



1+cscA=cscA(1)+1/sinA(sinA) ... Substitution (note: csc(A) = 1/sin(A))



1+cscA=cscA+sinA/sinA ... multiplication



1+cscA=cscA+1 ... division



1+cscA=1+cscA ... use the commutative property of addition



So the identity is confirmed.