First slide

Trigonometric Identities

Question

$\frac{2 + \sin^{6} θ + \cos^{6} θ}{\sin^{2} θ + \cos^{4} θ} =$

Moderate

A

1
B

2
C

3
D

4

Solution

$\begin{array}{l} \frac{2 + \sin^{6} θ + \cos^{6} θ}{\sin^{2} θ + \cos^{4} θ} = \frac{2 + {(\sin^{2} θ + \cos^{2} θ)}^{3} - 3 \sin^{2} θ \cos^{2} θ (\sin^{2} θ + \cos^{2} θ)}{\sin^{2} θ + \cos^{2} θ (1 - \sin^{2} θ)} \\ = \frac{2 + (1 - 3 \sin^{2} θ \cos^{2} θ)}{1 - \sin^{2} \cos^{2} θ} \\ = \frac{3 (1 - \sin^{2} θ \cos^{2} θ)}{(1 - \sin^{2} \cos^{2} θ)} \\ = 3 \end{array}$

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