Expert: Sherman D. - 5/31/2009

Question
Please Help me to prove this Identities.I'm having head here.
1.
tan x + tan y
------------- =  (tan x)(tan y)
cot x + cot y

2.
sec^2 X - sec^2 Y = tan^2 - tan^2 Y

3.
csc^2 X + sec^2 X = csc^2 X sec^ X

Those" X " are an alphabet not the sign of multiplicaion.

Thank you so much.Please make my day :-)

Answer
1.)
(tan(x) + tan(y))/(cot(x) + cot(y))

(tan(x) + tan(y))/((1/tan(x)) + (1/tan(y)))

(tan(x) + tan(y))/((tan(x) + tan(y))/(tan(x)tan(y)))

(tan(x) + tan(y)) * ((tan(x)tan(y))/(tan(x) + tan(y)))

as you can tell tan(x) + tan(y) cancels out

so this leaves you with tan(x)tan(y) as your answer.

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2.)
sec(x)^2 - sec(y)^2 = tan(x)^2 - tan(y)^2

sec(x)^2 = tan(x)^2 + 1

(tan(x)^2 + 1) - (tan(y)^2 + 1) = tan(x)^2 - tan(y)^2

tan(x)^2 + 1 - tan(y)^2 - 1 = tan(x)^2 - tan(y)^2

tan(x)^2 - tan(y)^2 = tan(x)^2 - tan(y)^2

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if your wondering how you get

tan(x)^2 + 1 out of sec(x)^2

sec(x)^2 = 1/cos(x)^2

tan(x)^2 + 1 =
(sin(x)/cos(x))^2 + 1 =
(sin(x)^2 + cos(x)^2)/cos(x)^2 =
1/cos(x)^2 =
sec(x)^2

also if your wondering how sin(x)^2 + cos(x)^2 = 1, its because

sin(x)^2 = 1 - cos(x)^2
cos(x)^2 = 1 - sin(x)^2

so either the cos(x)^2 or sin(x)^2 would cancel out, leaving you with 1 as the remainder.

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csc(x)^2 + sec(x)^2 = csc(x)^2sec(x)^2
(1/sin(x)^2) + (1/cos(x)^2) = (1/sin(x)^2)(1/cos(x)^2)
(cos(x)^2 + sin(x)^2)/(sin(x)^2cos(x)^2) = (1/(sin(x)^2cos(x)^2))
(1/(sin(x)^2cos(x)^2)) = (1/(sin(x)^2cos(x)^2))

by the way, to correct this problem, when your doing multiplication they use * and for the variable, they just use x, so

2x = 2*x and not 2*

info found at http://math2.org/math/trig/identities.htm