If 3 4 < < , then 2 +1 ^2 is equal to - Vidyadip

MCQ

If $\frac{3 \pi}{4}<\alpha<\pi$, then $\sqrt{2 \cot \alpha+\frac{1}{\sin ^2 \alpha}}$ is equal to

A
$-1+\cot \alpha$
✓
$-1-\cot \alpha$
C
$1-\cot \alpha$
D
$1+\cot \alpha$

✓

Answer

Correct option: B.

$-1-\cot \alpha$

We have:
$\sqrt{2 \cot \alpha+\frac{1}{\sin ^2 \alpha}}$
$=\sqrt{\frac{2 \cos \alpha}{\sin \alpha}+\frac{1}{\sin ^2 \alpha}}$
$=\sqrt{\frac{2 \sin \alpha \cos \alpha+1}{\sin ^2 \alpha}}$
$=\sqrt{\frac{2 \sin \alpha \cos \alpha+\sin ^2 \alpha+\cos ^2 \alpha}{\sin ^2 \alpha}}$
$=\sqrt{\frac{(\sin \alpha+\cos \alpha)^2}{\sin ^2 \alpha}}$
$=\sqrt{(1+\cot \alpha)^2}$
$=|1+\cot \alpha|$
$=-(1+\cot \alpha)\left[\text { When } \frac{3 \pi}{4}<\alpha<\pi, \cot \alpha<-1 \Rightarrow \cot \alpha+1<0\right]$
$=-1-\cot \alpha$

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