In general, tan (α + β) is not equal to tan α + tan β. However, there are some values of α and β for which they are equal. Find such α and β, and do t

Amari Flowers 2021-06-06 Answered

In general, \(\tan(\alpha+\beta)\) is not equal to \(\tan\alpha+\tan\beta\). However, there are some values of \(\alpha\) and \(\beta\) for which they are equal. Find such \(\alpha\) and \(\beta\) and do the same for \(\tan(\alpha-\beta)\)

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curwyrm
Answered 2021-06-07 Author has 25420 answers

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 \(\beta=0^{\circ}\)

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Answered 2021-08-11 Author has 11086 answers

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