Consider the left hand side,
cscA+cotA–1cotA−cscA+1 ------------(1)
We know that, csc2A−cot2A=1
Substituting this in the numerator of...............(1)
cscA+cotA−(csc2A−cot2A)(cotA−cscA+1)
We know that, x2−y2=(x+y)(x−y).
Applying it in the numerator,
cscA+cotA−(cscA+cotA)(cscA−cotA)(cotA−cscA+1)
Taking common,
=(cscA+cotA)(1−cscA+cotA)(cotA−cscA+1)
Canceling like terms in numerator and denominator, we are left with,
cscA+cotA
=1sinA+(cosAsinA)
=(1+cosA)sinA