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

User SkaveRat
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[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
User Brandyn
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