Question

$\tan(A+B)=m$ and $\tan(A-B)=n$. Then $\tan(2A)=$

• A. $\frac{m+n}{1+mn}$
• B. $\frac{m-n}{1+mn}$
• C. $\frac{m+n}{1-mn}$
• D. $\frac{m-n}{1-mn}$

Hint:

Whenever $(A+B)$ and $(A-B)$ appear together, add them to obtain $2A$ and apply standard trigonometric identities.

Answer 1

The Correct Option is C

Concept: Use the identity \[ \tan(X+Y)=\frac{\tan X+\tan Y}{1-\tan X\tan Y} \] and observe that \[ 2A=(A+B)+(A-B). \]

Step 1: Express $\tan(2A)$.
\[ \tan(2A) = \tan[(A+B)+(A-B)] \] Applying the tangent addition formula, \[ \tan(2A) = \frac{\tan(A+B)+\tan(A-B)} {1-\tan(A+B)\tan(A-B)} \]

Step 2: Substitute the given values.
Given, \[ \tan(A+B)=m,\qquad \tan(A-B)=n \] Therefore, \[ \tan(2A) = \frac{m+n}{1-mn} \]