• # question_answer If $\frac{3\pi }{4}<\alpha <\pi ,$then $\sqrt{\text{cose}{{\text{c}}^{2}}\alpha +2\cot \alpha }$ is equal to  [Pb. CET 2000; AMU 2001; MP PET 2004] A) $1+\cot \alpha$ B) $1-\cot \alpha$ C) $-1-\cot \alpha$ D) $-1+\cot \alpha$

$\sqrt{\text{cose}{{\text{c}}^{2}}\alpha +2\cot \alpha }=\sqrt{1+{{\cot }^{2}}\alpha +2\cot \alpha }=\,\,|1+\cot \alpha |$ But $\frac{3\pi }{4}<\alpha <\pi \Rightarrow \cot \alpha <-1\Rightarrow 1+\cot \alpha <0$ Hence, $|1+\cot \alpha |=-(1+\cot \alpha )$.