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