Question

If $\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\text{cosec}\theta +\cot \theta$, where $(\theta \in (0,2\pi )-\{\pi \left\}\right)$, then $\theta$ lies in

• A. first quadrant
• B. second quadrant
• C. third quadrant
• D. fourth quadrant

Answer 1

Answer: (A), (B)

The correct options are
A first quadrant
B second quadrant

$\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\text{cosec}\theta +\cot \theta$

$\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\sqrt{\frac{1+\cos \theta }{1-\cos \theta }\times \frac{1+\cos \theta }{1+\cos \theta }}=\sqrt{\frac{(1+\cos \theta {)}^2}{{\sin }^2\theta }}=\frac{|1+\cos \theta |}{|\sin \theta |}=\frac{1+\cos \theta }{|\sin \theta |}(\because 1+\cos \theta \ge 0)=\frac{1}{|\sin \theta |}+\frac{\cos \theta }{|\sin \theta |}$

When $\theta \in (0,\pi )$, then
$\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\text{cosec}\theta +\cot \theta$

When $\theta \in (\pi ,2\pi )$, then
$\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=-\text{cosec}\theta -\cot \theta$

Therefore, $\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\text{cosec}\theta +\cot \theta$ when $\theta$ is in first and second quadrant.