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

1 answer:

4 0

[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

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7 0

3 years ago

Pls help many thanks

salantis [7]

So, let's make these two values improper fractions so they're easier to work with. $4+\frac{1}{3}=\frac{12}{3}+\frac{1}{3}=\frac{13}{3} 3+\frac{1}{2}=\frac{6}{2}+\frac{1}{2}=\frac{7}{2}$