Question:
If $\frac{3 \pi}{4}<\alpha<\pi$, then $\sqrt{2 \cot \alpha+\frac{1}{\sin ^{2} \alpha}}$ is equal to
(a) 1 − cot α
(b) 1 + cot α
(c) −1 + cot α
(d) −1 −cot α
Solution:
(d) −1 −cot α
We have:
$\sqrt{2 \cot \alpha+\frac{1}{\sin ^{2} \alpha}}$
$=\sqrt{\frac{2 \cos \alpha}{\sin \alpha}+\frac{1}{\sin ^{2} \alpha}}$
$=\sqrt{\frac{2 \sin \alpha \cos \alpha+1}{\sin ^{2} \alpha}}$
$=\sqrt{\frac{2 \sin \alpha \cos \alpha+\sin ^{2} \alpha+\cos ^{2} \alpha}{\sin ^{2} \alpha}}$
$=\sqrt{\frac{(\sin \alpha+\cos \alpha)^{2}}{\sin ^{2} \alpha}}$
$=\sqrt{(1+\cot \alpha)^{2}}$
$=|1+\cot \alpha|$
$=-(1+\cot \alpha) \quad\left[\right.$ When $\left.\frac{3 \pi}{4}<\alpha<\pi, \cot \alpha<-1 \Rightarrow \cot \alpha+1<0\right]$
$=-1-\cot \alpha$