First slide

Trigonometric Identities

Question

If $π < θ < 2 π$ , then $\sqrt{\frac{1 + \cos θ}{1 - \cos θ}}$ is equal to

Moderate

A

$cosec θ + \cot θ$
B

$cosec θ - \cot θ$
C

$- cosec θ + \cot θ$
D

$- cosec θ - \cot θ$

Solution

we have,
$\begin{array}{l} \sqrt{\frac{1 + \cos θ}{1 - \cos θ}} = \sqrt{\frac{(1 + \cos θ)^{2}}{1 - \cos^{2} θ}} \\ \Rightarrow \sqrt{\frac{1 + \cos θ}{1 - \cos θ}} = \frac{1 + \cos θ}{| \sin θ |} \\ \Rightarrow \sqrt{\frac{1 + \cos θ}{1 - \cos θ}} = \frac{1 + \cos θ}{- \sin θ} [∵ π < θ < 2 π \Rightarrow \sin θ < 0] \end{array}$
$\Rightarrow \sqrt{\frac{1 + \cos θ}{1 - \cos θ}} = - cosec θ - \cot θ$

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