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Prove the identity.

cos(x - y) 1+ tan x tan y
------------ = ------------------
cos(x + y) 1- tan x tan y

here is the actual problem on the link below...thanks so much in advance for helping me.

http://www.fbixtreme.com/David/l1.jpg

2006-11-17 08:22:29 · 3 answers · asked by David N 1 in Science & Mathematics Mathematics

3 answers

okay,

cos(x-y) cos(x)*cos(y) + sin(x)*sin(y)
---------- = -----------------------------------
cos(x+y) cos(x)*cos(y) - sin(x)*sin(y)

(multiply and divide the fraction by cos(x)*cos(y))

1 + tan(x)*tan(y)
= ---------------------
1 - tan(x)*tan(y)

2006-11-17 09:51:40 · answer #1 · answered by farbod f 2 · 0 0

Find and use sum and difference formulas for cos and divide top and bottom by something to get that 1 out front.

2006-11-17 08:27:43 · answer #2 · answered by modulo_function 7 · 0 0

LHS = Tan^2 a / (a million + tan^2 a) Tan = sin/cos, so =[Sin^2(a)/Cos^2(a)]/[a million+Sin^2(a)/cos^2... placed the expression on the bottom over a complication-loose denominator = [sin^2(a)/cos^2(a)]/[(cos^2(a)+sin^2(a))... sin^2 + cos^2 = a million, so = [sin^2(a)/cos^2(a)]/[a million/cos^2(a)] = sin^2(a)/cos^2(a) * cos^2(a)/a million = sin^2(a) = RHS

2016-11-29 05:44:42 · answer #3 · answered by ? 4 · 0 0

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