A) \[1+\cot \alpha \]
B) \[1-\cot \alpha \]
C) \[-1-\cot \alpha \]
D) \[-1+\cot \alpha \]
Correct Answer: C
Solution :
\[\sqrt{\text{cose}{{\text{c}}^{2}}\alpha +2\cot \alpha }=\sqrt{1+{{\cot }^{2}}\alpha +2\cot \alpha }=\,\,|1+\cot \alpha |\] 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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