Take L.H.S
=>
SinA/cotA+cosecA
=sinA/(cosA/sinA+1/sinA)
=sinA/{(cosA+1)/sinA}
=sin²A/(1+cosA)
=(1-cos²A)/(1+cosA)
=(1+cosA)(1-cosA)/(1+cosA)
=1-cosA
Now take R.H.S
2+sinA/cotA-cosecA
=2+sinA/(cosA/sinA-1/sinA)
=2+sinA/{(cosA-1)/sinA}
=2+sin²A/(cosA-1)
=2+(1-cos²A)/{-(1-cosA)}
=2-(1+cosA)(1-cosA)/(1-cosA)
=2-(1+cosA)
=2-1-cosA
=1-cosA
∴, LHS=RHS (Proved)
I hope you understood the answer :)