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I just cant think at the mo, can someone prove-
tan(A+B)= tanA+tanB/1-tanAtanB

I know its a compound angle but how do u prove it equals that

2006-11-13 13:53:00 · 3 answers · asked by aye? 1 in Science & Mathematics Mathematics

3 answers

tan(a+b)=sin(a+b)/cos(a+b)=
(sinacosb+sinbcosa)/
(cosacosb-sinasinb)
divide top and bottom by cosacosb
(tana+tanb)/(1-tanatanb)

2006-11-13 13:56:36 · answer #1 · answered by Greg G 5 · 0 0

cos x /( a million+ sin x) - (a million-sin x) / cos x = 0 easy denominator is cosx(a million + sinx) so the equation turns into cos²x - (a million - sin²x) = 0 via Pythagorean id a million - sin²x = cos²x so cos²x - cos²x = 0 indefinite answer

2016-12-14 06:45:04 · answer #2 · answered by ? 4 · 0 0

tan(A+B)=sin(A+B)/cos(A+B)= (sinAcosB+sinBcosA) / (cosAcosB-sinAsinB)
= (tanA+tanB)/(1-tanAtanB) '

2006-11-13 14:18:25 · answer #3 · answered by Anonymous · 1 0

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