Question

Let $\frac{3\pi }{4}<\theta <\pi$ and $\sqrt{2\cot \theta +\frac{1}{{\sin }^2\theta }}=k-\cot \theta$, then $k$ is equal to

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$-1$

We have $\sqrt{2\cot \theta +\frac{1}{{\sin }^2\theta }}=k-{\cot }^{}\theta$

$\Rightarrow \sqrt{2\cot \theta +cose{c}^2\theta }=k-\cot \theta$

$\Rightarrow \sqrt{{(1+\cot \theta )}^2}=k-\cot \theta$

$\Rightarrow \pm (1+\cot \theta )=k-\cot \theta$

$\Rightarrow -(1+\cot \theta )=k-\cot \theta$

$\Rightarrow k=-1$