Question

If $\frac{3\mathrm{\pi }}{4}<\mathrm{\alpha }<\mathrm{\pi },\mathrm{then}\sqrt{2\cot \mathrm{\alpha }+\frac{1}{{\sin }^2\mathrm{\alpha }}}$ is equal to
(a) 1 − cot α
(b) 1 + cot α
(c) −1 + cot α
(d) −1 −cot α

Answer 1

(d) −1 −cot α

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \sqrt{2\mathrm{cot\alpha }+\frac{1}{{\sin }^2\mathrm{\alpha }}} \\ =\sqrt{\frac{2\cos \alpha }{\sin \alpha }+\frac{1}{{\sin }^2\mathrm{\alpha }}} \\ =\sqrt{\frac{2\sin \alpha \cos \alpha +1}{{\sin }^2\mathrm{\alpha }}} \\ =\sqrt{\frac{2\sin \alpha \cos \alpha +{\sin }^2\mathrm{\alpha }+{\cos }^2\mathrm{\alpha }}{{\sin }^2\mathrm{\alpha }}} \\ =\sqrt{\frac{{\left(\sin \alpha +\cos \alpha \right)}^2}{{\sin }^2\mathrm{\alpha }}} \\ =\sqrt{{\left(1+\cot \alpha \right)}^2} \\ =\left|1+\cot \alpha \right| \\ =-\left(1+\cot \alpha \right)\left[\mathrm{When}\frac{3\mathrm{\pi }}{4}<\mathrm{\alpha }<\mathrm{\pi },\cot \alpha <-1\Rightarrow \mathrm{c}\mathrm{o}\mathrm{t}\mathrm{\alpha }+1<0\right] \\ =-1-\mathrm{c}\mathrm{o}\mathrm{t}\alpha \end{gathered}$$