Question

If $\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\text{cosec}\theta +\cot \theta$, where $\theta =\frac{k\pi }{8},k\in N$, then the number of possible value(s) of $\theta \in [0,2\pi ]$ is

• A. 7.00
• B. 7.0
• C. 7

Answer 1

Answer: (A), (B), (C)

Given
$\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\text{cosec}\theta +\cot \theta$
So,
$\theta \ne \{0,\pi ,2\pi \}$
Now,
$\sqrt{\frac{1+\cos \theta }{1-\cos \theta }\times \frac{1+\cos \theta }{1+\cos \theta }}=\text{cosec}\theta +\cot \theta \Rightarrow \sqrt{\frac{(1+\cos \theta {)}^2}{1-{\cos }^2\theta }}=\text{cosec}\theta +\cot \theta \Rightarrow \frac{|1+\cos \theta |}{|\sin \theta |}=\text{cosec}\theta +\cot \theta \Rightarrow \frac{1+\cos \theta }{|\sin \theta |}=\text{cosec}\theta +\cot \theta (\because 1+\cos \theta \ge 0)\Rightarrow \frac{1+\cos \theta }{|\sin \theta |}=\frac{1+\cos \theta }{\sin \theta }$
It is possible if
$\sin \theta >0\Rightarrow \pi >\theta >0\Rightarrow \pi >\frac{k\pi }{8}>0\therefore k\in (0,8),k\in N$

Hence, there are $7$ possible values of $k$.