Question

If $\frac{3\pi }{4}<\alpha <\pi$, then $\sqrt{cose{c}^2\alpha +2cot\alpha }$ 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 C

$-1-cot\alpha$

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