If 3 4 < < , then cose c ^2 + 2 is equal to - Vidyadip

MCQ

If $\frac{{3\pi }}{4} < \alpha < \pi ,$ then $\sqrt {{\rm{cose}}{{\rm{c}}^2}\alpha + 2\cot \alpha } $ is equal to

A
$1 + \cot \alpha $
B
$1 - \cot \alpha $
✓
$ - 1 - \cot \alpha $
D
$ - 1 + \cot \alpha $

✓

Answer

Correct option: C.

$ - 1 - \cot \alpha $

c
(c) $\sqrt {{\rm{cose}}{{\rm{c}}^2}\alpha + 2\cot \alpha } $

$= \sqrt {1 + {{\cot }^2}\alpha + 2\cot \alpha } = \,\,|1 + \cot \alpha |$

But $\frac{{3\pi }}{4} < \alpha < \pi \Rightarrow \cot \alpha < - 1 $

$\Rightarrow 1 + \cot \alpha < 0$

Hence, $|1 + \cot \alpha | = - (1 + \cot \alpha )$.

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