If frac{{3pi }}{4}

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Question

(c) $\sqrt {{\rm{cose}}{{\rm{c}}^2}\alpha + 2\cot \alpha } $

$= \sqrt {1 + {{\cot }^2}\alpha + 2\cot \alpha } = \,\,|1 + \cot \alpha |$ 

Bu

Vedclass · Accepted Answer

$ - 1 - \cot \alpha $

Vedclass · Answer

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

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

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