Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

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