First slide

Trigonometric Identities

Question

If $\frac{3 π}{4} < α < π, then \sqrt{2 \cot α + \frac{1}{\sin^{2} α}}$ is equal to

Moderate

A

$1 + \cot α$
B

$- 1 - \cot α$
C

$1 - \cot α$
D

$- 1 + \cot α$

Solution

$\begin{matrix} \sqrt{2 \cot α + \frac{1}{\sin^{2} α}} = \sqrt{2 \cot α + {cosec}^{2} α} \\ = \sqrt{2 \cot α + 1 + \cot^{2} α} \\ = | 1 + \cot α | = - 1 - \cot α \\ [since \cot α < - 1 when 3 π / 4 < α < π] \end{matrix}$

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