Question

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

• A. $1+\cot \alpha$
• B. $-1-\cot \alpha$
• C. $1-\cot \alpha$
• D. $-1+\cot \alpha$

Answer 1

Answer: (B)

$-1-\cot \alpha$

Let $y=\sqrt{2cot\alpha +\frac{1}{si{n}^2\alpha }}$
$=\sqrt{2\frac{cos\alpha }{sin\alpha }+\frac{1}{si{n}^2\alpha }}=\sqrt{\frac{1+2sin\alpha .cos\alpha }{si{n}^2\alpha }}$
$=\sqrt{\frac{(cos\alpha +sin\alpha {)}^2}{si{n}^2\alpha }}=\mid \frac{(cos\alpha +sin\alpha )}{sin\alpha }\mid$

$=\mid 1+cot\alpha \mid$

but given $\frac{3\pi }{4}<\alpha <\pi \Rightarrow -\infty <cot\alpha <-1$

Hence, $y=-1-cot\alpha$