Question

# Starting with the addition formulas for the sine and cosine, derive the addition formula$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$for the tangent.

Solution

Verified

We will be using the addition formulas for Sine and Cosine which are,

\begin{align*} \sin(A+B)&=\sin A\cos B+\sin B\cos A \\ \cos(A+B)&=\cos A\cos B-\sin A\sin B \\ \end{align*}

Now computing the derivation of $\tan(A+B)$,

\begin{align*} \tan(A+B)&=\frac{\sin(A+B)}{\cos(A+B)} &&\left(\text{By} \tan\theta=\frac{\sin\theta}{\cos\theta}\right)\\ &=\frac{\sin A\cos B+\sin B\cos A}{\cos A\cos B-\sin A\sin B} &&(\text{By addition formulas}) \\ \end{align*}

Now dividing numerator and denominator by $\cos A\cos B$,

\begin{align*} \tan(A+B)&=\frac{\dfrac{\sin A\cos B+\sin B\cos A}{\cos A\cos B}}{\dfrac{\cos A\cos B-\sin A\sin B}{\cos A\cos B}} \\ &=\frac{\dfrac{\sin A\cos B}{\cos A\cos B}+\dfrac{\sin B\cos A}{\cos A\cos B}}{\dfrac{\cos A\cos B}{\cos A\cos B}-\dfrac{\sin A\sin B}{\cos A\cos B}} \\ &=\frac{\dfrac{\sin A}{\cos A}+\dfrac{\sin B}{\cos B}}{1-\dfrac{\sin A\sin B}{\cos A\cos B}} \\ &=\frac{\tan A+\tan B}{1-\tan A\tan B}&&\left(\text{By}\tan\theta=\frac{\sin\theta}{\cos\theta}\right) \end{align*}

Hence derived.