Chapter 7.2, Problem 71E

To determine

To prove: the addition formula for tangent.

Explanation of Solution

The addition formula for the tangent is,

  $\tan \left(s+t\right)=\frac{\tan s+\tan t}{1-\tan s\tan t}$

To prove this formula, simplify the left hand side and try to get the right hand side.

The left hand side as,

  $L.H.S=\tan \left(s+t\right)$

On using $\tan \theta =\frac{\sin \theta }{\cos \theta }$ , get

  $L.H.S=\frac{\sin \left(s+t\right)}{\cos \left(s+t\right)}$

ON using the addition of the formulas for the Cosine as,

  $\cos \left(s+t\right)=\cos s\cos t-\sin s\sin t$ and the addition of the formulas for the Sine as,

  $\sin \left(s+t\right)=\sin s\cos t+\cos s\sin t$ , get

  $L.H.S=\frac{\sin s\cos t+\cos s\sin t}{\cos s\cos t-\sin s\sin t}$

ON dividing the numerator and the denominator by $\cos s\cos t$ , get

  $\begin{array}{c}L.H.S=\frac{\frac{\sin s\cos t+\cos s\sin t}{\cos s\cos t}}{\frac{\cos s\cos t-\sin s\sin t}{\cos s\cos t}}\\ =\frac{\frac{\sin s\cos t}{\cos s\cos t}+\frac{\cos s\sin t}{\cos s\cos t}}{\frac{\cos s\cos t}{\cos s\cos t}-\frac{\sin s\sin t}{\cos s\cos t}}\\ =\frac{\frac{\sin s}{\cos s}+\frac{\sin t}{\cos t}}{1-\frac{\sin s}{\cos s}\times \frac{\sin t}{\cos t}}\end{array}$

On using $\tan \theta =\frac{\sin \theta }{\cos \theta }$ , get

  $\begin{array}{c}L.H.S=\frac{\tan s+\tan t}{1-\tan s\tan t}\\ =R.H.S\end{array}$

Thus, get

  $\tan \left(s+t\right)=\frac{\tan s+\tan t}{1-\tan s\tan t}$ .