Answer
$\tan (s+t)=\dfrac{\tan s+\tan t}{1-\tan s \tan t}$
Work Step by Step
Use the identities $\cos (s+t)=\cos s \cos t -\sin s \sin t$
and $\sin (s+t)=\sin s \cos t +\cos s \sin t$
$\tan (s+t)=\dfrac{\sin (s+t)}{\cos (s+t)}$
This gives:
$\tan (s+t)=\dfrac{\sin s \cos t +\cos s \sin t}{\cos s \cos t -\sin s \sin t$}$
or, $\tan (s+t)=\dfrac{\dfrac{\sin s}{\cos s}+\dfrac{\sin t}{\cos t}}{1-(\dfrac{\sin s}{\cos s})(\dfrac{\sin t}{\cos t})}$
Thus,
$\tan (s+t)=\dfrac{\tan s+\tan t}{1-\tan s \tan t}$
