First slide

Compound angles

Question

$\frac{3 + \cot 76 ° \cot 16 °}{\cot 76 ° + \cot 16 °}$ is equal to

Moderate

A

$\tan 16 °$
B

$\tan 76 °$
C

$\tan 44 °$
D

$\tan 46 °$

Solution

$\begin{array}{l} \frac{3 + \cot 76 ° \cos 16 °}{\cot 76 ° + \cot 16 °} \\ = \frac{3 \sin 76 ° \sin 16 ° + \cos 76 ° \cos 16 °}{\sin (76 ° + 16 °)} \\ = \frac{2 \sin 76 ° \sin 16 ° + \cos (76 ° - 16 °)}{\sin 92 °} \\ = \frac{\cos 60 ° - \cos 92 ° + \cos 60 °}{\sin 92 °} \\ = \frac{1 - \cos 92 °}{\sin 92 °} \\ = \frac{2 \sin^{2} 46^{°}}{2 \sin 46 ° \cos 46 °} \\ = \tan 46 ° = \tan (90 ° - 44 °) = \cot 44 ° \end{array}$

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