Question

If $\frac{3\pi }{4}<\alpha <\pi$, then $\sqrt{2\cot \alpha +\frac{1}{{\sin }^2\alpha }}$ is equal to

• A. $1+\cot \alpha$
• B. $-1-\cot \alpha$
• C. $1-\cot \alpha$
• D. $-1+\cot \alpha$

Answer 1

The correct option is B $-1-\cot \alpha$

$\sqrt{2\cot \alpha +\frac{1}{{\sin }^2\alpha }}$

$=\sqrt{2\cot \alpha +{\text{cosec}}^2\alpha }$

$=\sqrt{2\cot \alpha +1+{\cot }^2\alpha }$

$=|1+\cot \alpha |$

$=-1-\cot \alpha [\because \cot \alpha <-1\text{when}\frac{3\pi }{4}<\alpha <\pi ]$