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3.Trigonometrical Ratios, Functions and Identities
easy
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 $
C
$ - 1 - \cot \alpha $
D
$ - 1 + \cot \alpha $
Solution
(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, $|1 + \cot \alpha | = – (1 + \cot \alpha )$.
Standard 11
Mathematics
