First slide

Trigonometric Identities

Question

The set of all possible values of $α$ in $[- π, π]$ such that $\sqrt{\frac{1 - \sin α}{1 + \sin α}}$ is equal to $\sec α - \tan α$ , is

Moderate

A

$[0, π / 2)$
B

$[0, π / 2) \cup (π / 2, π]$
C

$[- π, 0]$
D

$(- π / 2, π / 2)$

Solution

Clearly, $\sec α - \tan α$ is not defined for $α = \pm π / 2$ Now,
$\begin{aligned} \sqrt{\frac{1 - \sin α}{1 + \sin α}} = \sqrt{\frac{(1 - \sin α)^{2}}{\cos^{2} α}} \\ \Rightarrow & \sqrt{\frac{1 - \sin α}{1 + \sin α}} = \frac{1 - \sin α}{| \cos α |} = \sec α - \tan α, if \cos α > 0 \end{aligned}$
Clearly, $\cos α > 0 \Rightarrow α \in (- π / 2, π / 2)$

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