Given that sinA=3/5,90deg<A<180deg,and cosB= -5/13,180<B<270,find the value of tan (A+B).?

Question

can't use the calculator.If possible,show me the steps.I'm really not good in trig function.

Answer 1

tan A = sinA/cosA= sinA /sqrt(1-(sinA)^2) or (3/5)/sqrt(1-9/25)=-3/4 Because A lies in 2nd quadrant
tan B = sinB/cosB = sqrt((1-(cosB)^2))/cosB or sqrt(144/169)/(-5/13) = 12/5
as sin B has to be - because B is in 3rd quadrant.
tan(A+B) =
(tanA+tanB)/(1-tanAtanB)
=33/20/(1+36/20)=33/56

Answer 2

Tan (α+β) = (Tanα + Tanβ)/(1-(Tanα*Tanβ))

That is a really useful formula. The formula for Tangent is:

Tan α = Sinα / Cosα

For 90º < α < 180º, due to the CAST rule, the Sine is a positive value in that range. Cosine is a negative value in there. Therefore, you should end up with a negative value (because of Cosine being negative).

Now plug in the values and you will get the result.

Good luck.

Answer 3

tan (A + B)
= (tan A + tan B) / (1 - tan A tan B)
= [ (- 3/4) + (12/5) ] / [ (1 - (- 3/4)(12/5) ]
= ( 33 / 20) / [ 1 + 36 / 20 ]
= (33 / 20) / (56 / 20)
= 33 / 56

Answer 4

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