Answer
$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
Work Step by Step
Derive a formula for $\tan(A-B)$
- First, use the identity: $\tan\theta=\frac{\sin\theta}{\cos\theta}$
$$\tan(A-B)=\frac{\sin(A-B)}{\cos(A-B)}$$
- Apply the addition formulas:
$$\tan(A-B)=\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B+\sin A\sin B}$$
- Divide both the numerator and denominator by $\cos A\cos B$:
$$\tan(A-B)=\frac{\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}}{\frac{\cos A\cos B+\sin A\sin B}{\cos A\cos B}}$$
$$\tan(A-B)=\frac{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}{1+\frac{\sin A\sin B}{\cos A\cos B}}$$
- Now we can replace $\frac{\sin A}{\cos A}$ with $\tan A$ and $\frac{\sin B}{\cos B}$ with $\tan B$.
$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
We have finished deriving a formula for $\tan(A-B)$
