Question

\[ \text{If } \frac{3\pi}{4}<\alpha<\pi, \]

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

is equal to

(a) \(1-\cot\alpha\)
(b) \(1+\cot\alpha\)
(c) \(-1+\cot\alpha\)
(d) \(-1-\cot\alpha\)

Solution

\[ \sqrt{2\cot\alpha+\csc^2\alpha} \]

Using identity

\[ \csc^2\alpha=1+\cot^2\alpha \]

\[ = \sqrt{2\cot\alpha+1+\cot^2\alpha} \]

\[ = \sqrt{(1+\cot\alpha)^2} \]

\[ =|1+\cot\alpha| \]

Since

\[ \frac{3\pi}{4}<\alpha<\pi \]

\(\alpha\) lies in II quadrant and

\[ \cot\alpha<-1 \]

Therefore,

\[ 1+\cot\alpha<0 \]

\[ |1+\cot\alpha| =-(1+\cot\alpha) =-1-\cot\alpha \]

Answer

\[ \boxed{-1-\cot\alpha} \]

Correct Option: (d)