Chapter 5.3, Problem 62E

Solution

To prove: The identity $\tan \left(u+v\right)=\frac{\tan u+\tan v}{1-\tan u\cdot \tan v}$ .

It is proved that $\tan \left(u+v\right)=\frac{\tan u+\tan v}{1-\tan u\cdot \tan v}$ .

Given information:

The given identity is $\tan \left(u+v\right)=\frac{\tan u+\tan v}{1-\tan u\cdot \tan v}$ .

Formula used:

Use the formula for sine of sum.

  $\sin \left(u+v\right)=\sin u\cos v+\cos u\sin v$

Use the formula for cosine of sum.

  $\cos \left(u+v\right)=\cos u\cos v-\sin u\sin v$

Proof:

Rewrite the left hand side of the given identity by using Reciprocal identity.

  $\begin{array}{c}\tan \left(u+v\right)=\frac{\sin \left(u+v\right)}{\cos \left(u+v\right)}\\ =\frac{\sin u\cos v+\cos u\sin v}{\cos u\cos v-\sin u\sin v}\end{array}$

Divide numerator and denominator by $\cos u\cos v$ .

  $\begin{array}{c}\tan \left(u+v\right)=\frac{\frac{\sin u\cos v+\cos u\sin v}{\cos u\cos v}}{\frac{\cos u\cos v-\sin u\sin v}{\cos u\cos v}}\\ =\frac{\frac{\sin u\cos v}{\cos u\cos v}+\frac{\cos u\sin v}{\cos u\cos v}}{\frac{\cos u\cos v}{\cos u\cos v}-\frac{\sin u\sin v}{\cos u\cos v}}\\ =\frac{\frac{\sin u}{\cos u}+\frac{\sin v}{\cos v}}{1-\frac{\sin u}{\cos u}\cdot \frac{\sin v}{\cos v}}\\ =\frac{\tan u+\tan v}{1-\tan u\tan v}\end{array}$

Hence, it is proved that $\tan \left(u+v\right)=\frac{\tan u+\tan v}{1-\tan u\cdot \tan v}$ .