Question

If $3\pi /4<\alpha <\pi$, 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

Answer: (B)

$-1-\cot \alpha$

$\sqrt{2\cot \alpha +\frac{1}{{\sin }^2\alpha }}$
$=\sqrt{2\cot \alpha +{\text{cosec}}^2\alpha }$
$=\sqrt{1+2\cot \alpha +{\cot }^2\alpha }$
$=\sqrt{(1+\cot \alpha {)}^2}$
$=|1+\cot \alpha |$.
Since $\alpha$ is in the second quadrant and $\frac{3\pi }{4}<\alpha <\pi$.
Hence, $|1+\cot \alpha |$$=-1-\cot \alpha$.