Question

If tan(A+B)=3/4 and tan(A-B)=77/36 then find tan2A​

Answer 1

Here, we will use the formula:

tan (C+D) = (tan C + tan D)/(1- tan C × tan D)

Let C = A+B

D = A-B

Then C+D = A+B+A-B = 2A

tan 2A = (3/4 + 77/36)/(1 - 3/4 × 77/36)

= (104/36)/(-29/48)

= 104/36 × -48/29

= -416/87

Answer 2

Given,

tan(A+B) = 3/4 and tan(A-B)=77/36

To find,

tan2A

Solution,

We know that,

Let (A + B) = C  and tan (A - B) = D

Then, tan C = 3/4 and tan D = 77/36.

We know that,

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

Similarly,

⇒   tan(C+D) = (tan C +tan D) / (1 - tan C tan D)

⇒   tan(C + D) = $\frac{3}{4}+\frac{77}{36}$ / $1-\frac{3}{4}*\frac{77}{36}$

⇒   tan(C + D) = $\frac{27+77}{36}$ / $\frac{48-77}{48}$

⇒   tan (C + D) = $\frac{104}{36} / \frac{-29}{48}$

⇒   tan (C + D) = -416/87

Now put C = A+B and D  =A-B

tan(C + D) = tan ( A+B + A - B) = tan 2A

Hence, tan 2A = -416/87.