Question

Let $\frac{3\pi }{4}<\theta <\pi$ and $\sqrt{2\cot \theta +\frac{1}{{\sin }^2\theta }}=a+b\cot \theta$ then $a+b=$ ?

• A. $2$
• B. $0$
• C. $-2$
• D. $-3$

Answer 1

Answer: (C)

$-2$

$\sqrt{2\cot \theta +\frac{1}{{\sin }^2\theta }}$
$=\sqrt{2\cot \theta +{\csc }^2\theta }$
$=\sqrt{2\cot \theta +1+{\cot }^2\theta }$
$=\sqrt{(1+\cot \theta {)}^2}$
$=|1+\cot \theta |$
For $\frac{3\pi }{4}<\theta <\pi$, $1+\cot \theta <0$
$\therefore |1+\cot \theta |=-1-\cot \theta$
$\therefore a=-1,b=-1⟹a+b=-2$