First slide

Compound angles

Question

${(\frac{\tan 70^{o} - \tan 20^{o}}{2 \tan 50^{o}} - \frac{\tan 50^{o} - \tan 40^{o}}{2 \tan 10^{o}})}^{2020} =$

Easy

A

0
B

2
C

-1
D

1

Solution

$\begin{array}{l} If A + B = 90 °, t h e n \tan A - \tan B = 2 \tan (A - B) \\ ∴ \frac{\tan 70 ° - \tan 20 °}{2 \tan 50 °} = \frac{2 \tan 50 °}{2 \tan 50 °} = 1 and \frac{\tan 50 ° - \tan 40 °}{2 \tan 10 °} = \frac{2 \tan 10 °}{2 \tan 10 °} = 1 \end{array} ∴ {(\frac{\tan 70 ° - \tan 20 °}{2 \tan 50 °} - \frac{\tan 50 ° - \tan 40 °}{2 \tan 10 °})}^{2020} = {(1 - 1)}^{2020} = 0$

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