Answer
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
- Start with the right side first.
- Use the identity: $\tan\theta=\frac{\sin\theta}{\cos\theta}$ to replace tangent functions.
- Simplify.
- Then use the addition formulas and simplify again.
Work Step by Step
Prove the identity:
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
We start with the right side first:
$$\frac{\tan A+\tan B}{1-\tan A\tan B}$$
- Recall the identity: $\tan\theta=\frac{\sin\theta}{\cos\theta}$
$$\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{1-\frac{\sin A}{\cos A}\times\frac{\sin B}{\cos B}}$$
$$\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B}}{1-\frac{\sin A\sin B}{\cos A\cos B}}$$
$$\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B}}{\frac{\cos A\cos B-\sin A\sin B}{\cos A\cos B}}$$
$$\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}$$
- Here, recall the addition formulas:
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
$$\cos(A+B)=\cos A\cos B-\sin A\sin B$$
Therefore, $$\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\sin(A+B)}{\cos(A+B)}$$
$$\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan(A+B)$$
The given identity has been proved.
