"How do you simplify and evaluate tan⁡80∘−tan⁡20∘1+tan⁡80∘tan⁡20∘" was answered by our experts

gabolzm6d

Answered question

2022-04-12

How do you simplify and evaluate

\frac{{\tan 80}^{\circ} - {\tan 20}^{\circ}}{1 + {\tan 80}^{\circ} {\tan 20}^{\circ}}

chambasos6

Beginner2022-04-13Added 12 answers

Method 1: Notice formula,

\frac{\tan A - \tan B}{1 + \tan A \tan B} = \tan (A - B)

Hence, we have

\frac{{\tan 80}^{\circ} - {\tan 20}^{\circ}}{1 + {\tan 80}^{\circ} {\tan 20}^{\circ}}

= \tan (80^{\circ} - 20^{\circ})

= \tan (60^{\circ})

= \sqrt{3}

cm1mmeboulbes21e1

Beginner2022-04-14Added 8 answers

Method 2: Notice,

\sin A \cos B - \cos A \sin B = \sin (A - B)

and

\cos A \cos B + \sin A \sin B = \cos (A - B)

Now, we have

\frac{{\tan 80}^{\circ} - {\tan 20}^{\circ}}{1 + {\tan 80}^{\circ} {\tan 20}^{\circ}}

= \frac{\frac{{\sin 80}^{\circ}}{{\cos 80}^{\circ}} - \frac{{\sin 20}^{\circ}}{{\cos 20}^{\circ}}}{1 + \frac{{\sin 80}^{\circ}}{{\cos 80}^{\circ}}}

= \frac{\sin (80^{\circ} - 20^{\circ})}{\cos (80^{\circ} - 20^{\circ})}

= \frac{{\sin 60}^{\circ}}{{\cos 60}^{\circ}}

= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}

= \sqrt{3}

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