some trigonometry problem?

Question

please take a look at the link below
http://img101.imagevenue.com/img.php?image=91337_DSC01619_122_907lo.JPG
how can I derive the following fomula according to the one from the link?
Tan(A+B)=tanA+tanB/1-tanATanB
I just couldn't figure out the derivation

by the way, how can I also convert 3+square root3/3-square root3 from sqr6+sqr2/sqr6-sqr2?
thanks for you. pal

Answer 1

Tan(a) = Sin(a) / Cos(a)

So the left side is:
Sin(a+b) / Cos(a+b)
= Tan(a+b)

For the right side, you divide every term by Cos(a)Cos(b)
This gives: (Sin(a)/Cos(a) + Sin(b)/Cos(b)) / (1 - Sin(a)/Cos(a) * Sin(b)/Cos(b))
=(Tan(a) + Tan(b)) / (1 - Tan(a)Tan(b)

Thus we have both sides:
Tan(a+b) = (Tan(a) + Tan(b)) / (1 - Tan(a)Tan(b)



For sqrt6+sqrt2 / sqrt6-sqrt2
You can divide the top and bottom by sqrt2, giving:
(sqrt3 + sqrt1) / (sqrt3 - sqrt1)
= (sqrt3 + 1) / (sqrt3 - 1)
... Maybe your answer is wrong?

Answer 2

sin (A + B) = sin A.cos B + cos A.sin B
cos (A + B) = cos A.cos B - sin A.sin B
Divide top and bottom of RHS by cos A.cos B
tan (A + B) = (tan A + tan B) / (1 - tan A.tan B)

Assume question is:-
(√6 + √2) / (√6 - √2) - (3 + √3) / (3 - √3)
Rationalise denominator of each fraction
= (8 + 2√2) / 4 - (12 + 6.√3) / 6
= (1/2).√2 - √3

Answer 3

sin(A+B) = sinAcosB + sinBcosA
cos(A+B) = cosAcosB - sinBsinA

tan(A+B) = [sinAcosB + sinBcosA] / [cosAcosB - sinBsinA]
Now divide every term by cosA cosB
tan(A+B) = [tanA + tanB] / (1 - tanAtanB)

Answer 4

divide each term on the R.H.S with cosAcosB