Let \(\displaystyleβ={0}°\) so that:

\(\tan(\alpha+0^{\circ})=\tan\alpha+\tan0^{\circ}\)

\(\tan\alpha=\tan\alpha\)

Hence,

\(\tan(\alpha+\beta)=\tan\alpha+\tan\beta\)

for any value of \(\alpha\alpha\) given that \(\beta=0^{\circ}\)

A possible answer is when:

\(\alpha=30^{\circ}\) and \(\beta=0^{\circ}\)

Do the same for \(\tan(\alpha−\beta)\). Let \(\beta=0^{\circ}\) so that:

\(\tan(\alpha−0^{\circ})=\tan\alpha−\tan0^{\circ}\)

\(\tan\alpha=\tan\alpha\)

Hence, \(\tan(\alpha-\beta)=\tan\alpha-\tan\beta\) for any value of αα given that \(\beta=0^{\circ}\). A possible answer is when:

\(\alpha=45^{\circ}\) and \(\beta=0^{\circ}\)