tan(A+B)=5 ,tan(A-B)=3 FIND tan2A & tan3A?

Answer 1

Here's a hint. Use the formula for expanding tan(X + Y) with X = A + B and Y = A - B. This should get you tan2A fairly easily. I think that to get tan3A you will first need to find tanA using the double angle formula in reverse but there might be a faster way. (I haven't spent very long trying to finish it.) Hope this helps.

Here's some serious help - ignore the previous answer and I don't agree with the quadratic in the middle of the next one either. Also I don't think that you will be allowed to get away with an approximate answer using trig tables or a calculator as shown later. I'm sure that you are required to find the exact answer as a radical if necessary.

Answer 2

tan(2A) = tan(A+B) + tan(A-B)
= {tan(A+B) + tan(A-B)} / {1 - tan(A+B)tan(A-B)}
= (5 + 3) / (1 - 5*3) = 8/(-14) = -4/7
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sin(2A) = 4/√65
cos(2A) = -7/√65

sin(A) = √[(1 - cos(2A))/2] = √[(1 + 7/√65)/2]
cos(A) = √[(1 + cos(2A))/2] = √[(1 - 7/√65)/2]

tan(A) = sin(A)/cos(A) = {√[(1 + 7/√65)/2]} / {√[(1 - 7/√65)/2]}
= √{(1 + 7/√65) / (1 - 7/√65)} = (1 + 7/√65) / √(1 - 49/65)
= (1 + 7/√65) / √(16/65) = (1 + 7/√65)√(16/65) / (16/65)
= [4/(√65) + 28/65] / (16/65) = (4√65 + 28) / 16
= (√65 + 7) / 4 ≈ 3.7655644
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tan(3A) = tan(2A + A)
= {-4/7 + (√65 + 7)/4} / {1 + (4/7)[(√65 + 7)/4]}
= [33/28 + (√65)/4] / {1 + (√65)/7 + 1)}
= [33/28 + (√65)/4] / [2 + (√65)/7]
= (33 + 7√65) / (56 + 4√65)
= (33*56 - 33*4√65 + 56*7√65 - 28*65) / (56² - 16*65)
= (28 + 260√65) / 2096
= (7 + 65√65) / 524 ≈ 1.013448

Answer 3

tan 2a=8
tan 3a=12

Answer 4

Tan (A+B)=5
A+B=inverse tan 5 = 78.69 ----------------equation 1

Tan (A-B)=3
A-B=inverse tan 3 = 71.57 ---------- equation 2

Adding equation 1 and 2

2A=150.26

Tan2A= Tan 150.26 = -0.57

Tan3A= Tan (150*3/2) = 1.01

Answer 5

Let tan(A)=x and tan(B)=y

tan(A+B)=(tan(a)+tan(B))/(1-tan(A)tan(B))
or

=(x+y)/(1-xy)=5

Similarly

tan(A-B)=(tan(A)-tan(B))/(1+tan(A)tan(B))

or (x-y)/(1+xy)=3

Through much manipulation you get 4x^2+x-19=0

so x=(-1+8sqrt(5))/8 and (-1-8sqrt(5))/8

so tan(A)=x

Next tan(2A)=(2tan(A)/(1-(tanA)^2) plug the value for x above

tan(3A)=tan(2A+A)=(tan(2A)+tan(A))/(1-tan(2A)tan(A)) we have all these values calculated, plug in.