Tan(y)+ cot(y)/csc(y)= sec(y) prove identity

[tan(y) + cot (y)]/csc (y)

tan (y) = sin (y)/cos (y)
cot (y) = cos (y)/sin (y)
csc (y) = 1/sin (y).
Now rewrite the expression with the equivalent values

[sin (y)/cos (y) + cos (y)/sin (y) ] / [1/sin (y)]

1st, let's work the Numerator only = [sin (y)/cos (y) + cos (y)].
                  = [(cos² (y) + sin² (y)]/ [cos (y).sin(y)]

or (cos² (y) + sin² (y) = 1, →Numerator = 1/[cos (y).sin(y)]

Denominator = csc (y) = [1/sin (y)], Then:

N/D = 1/[cos (y).sin(y)] / [1/sin (y)] =  [1 x sin (y)]/ [cos (y).sin (y)] = 1/cos (y)

 Or 1/cos (y) = sec (y) Q.E.D 

Ithink is a. true, but not sure