MCQ
If $\frac{3\pi}{4}<\text{a}<\pi,$ then $\sqrt{2\cot\text{a}+\frac{1}{\sin^\text{a}}}$ is equal to:
  • A
    $1-\cot\text{a}$
  • B
    $1+\cot\text{a}$
  • C
    $-1+\cot\text{a}$
  • $-1-\cot\text{a}$

Answer

Correct option: D.
$-1-\cot\text{a}$
We have:
$\sqrt{2\cot\alpha+\frac{1}{\sin\alpha}}$
$=\sqrt{\frac{2\cos\alpha}{\sin\alpha}+\frac{1}{\sin^2\alpha}}$
$=\sqrt{\frac{2\sin\alpha\cos\alpha+1}{\sin\alpha}}$
$=\sqrt{\frac{2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha}{\sin\alpha}}$
$=\sqrt{\frac{(\sin\alpha+\cos\alpha)^2}{\sin^2\alpha}}$
$=\sqrt{(1+\cot\alpha)^2}$
$=|1+\cot\alpha|$
$=-(1+\cot\alpha) $ $$ $[\text{ when}\frac{3\pi}{4}<\alpha<\pi,\cot\alpha<-1\Rightarrow\cot\alpha+1<0\big]$
$=-1-\cot\alpha$

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