Question

if tanA=1/2 and tanB=1/3, then value of tan2A+B is​

Answer 1

tan(A + B) = [(tanA + tanB)/1 - tanAtanB]

So,

tan(2A + B) = [(tan2A +tanB)/1 - tan2AtanB]

Also,

tan2A = 2tanA/(1 – tan²A)

⇒ tan2A = 2(1/2)/(1 – (1/2)²)

⇒ tan2A = 4/3

⇒ tan(2A + B) = [(4/3 + 1/3)/1 - (4/3)(1/3)]

⇒ tan(2A + B) = (5/3)/(5/9)

⇒ tan(2A + B) = 3

$\boxed{\displaystyle\sf{tan(2A + B) = 3}}$

Answer 2

Answer:

Step-by-step explanation:

Since we know by the double angle formulae that tan2X = (2tanX)/(1−tan^2X), we see as follows:

tan2(A+B)

= 2tan(A+B)/(1-tan^2(A+B))

Now, if we can determine what tan(A+B) equals, our life would be much easier!

Well, since tan(A + B) = sin(A+B)/cos(A+B), we can use the sum formulas of sin(A+B) and cos(A+B) to yield that tan(A+B) equals:

(AND THIS DESERVES A DRUMROLL!!!)

(tanA + tanB)/(1-tanAtanB)

Well, we know what tanA and tanB are, so substitute. You get, then, that tan(A+B) actually equals 1! Which means that A+B is 45 degrees (or pi/4 radians)

Substituting once more with our previous equation, then, we see that tan2(A+B) equals:

2(1)/(1-tan(45)^2)

= 2/(1-1)

= 2/0

which is undefined.