First slide

Trigonometric Identities

Question

If $x = \frac{2 \sin θ}{1 + \cos θ + \sin θ}, then \frac{1 - \cos θ + \sin θ}{1 + \sin θ}$ is equal to

Moderate

A

1+x
B

1-x
C

x
D

1/x

Solution

Multiplying the numerator and denominator of x by
$1 - \cos θ + \sin θ$ , we get
$\begin{matrix} x = \frac{2 \sin θ (1 - \cos θ + \sin θ)}{(1 + \sin θ)^{2} - \cos^{2} θ} \\ = \frac{2 \sin θ (1 - \cos θ + \sin θ)}{(1 + \sin θ)^{2} - (1 - \sin^{2} θ)} \\ = \frac{2 \sin θ \cdot (1 - \cos θ + \sin θ)}{(1 + \sin θ) (2 \sin θ)} \\ = \frac{1 - \cos θ + \sin θ}{1 + \sin θ} \end{matrix}$

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