First slide

Trigonometric Identities

Question

If $0 < θ < \frac{π}{2}$ then $\frac{\tan θ + \sec θ - 1}{\tan θ - \sec θ + 1}$ is equal to

Moderate

A

$1 + \sin θ + \cos θ$
B

$\frac{1 + \sin θ}{\cos θ}$
C

$\frac{1 - \cos θ}{\sin θ}$
D

$\tan θ - \sec θ$

Solution

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

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