The value of $\sqrt{\large\frac{1+\cos\theta}{1-\cos\theta}}$ is equal to

Question

$\begin{array}{1 1}(a)\;\left\{\begin{array}{1 1}-cosec\theta-\cot\theta&if\;0<\theta>\pi\\-cosec\theta-\cot\theta&if\;\pi<\theta<2\pi\end{array}\right.\\(b)\;\left\{\begin{array}{1 1}cosec\theta+\cot\theta&if\;0<\theta>\pi\\-cosec\theta+\cot\theta&if\;\pi<\theta<2\pi\end{array}\right.\\(c)\;\left\{\begin{array}{1 1}cosec\theta+\cot\theta&if\;0<\theta>\pi\\-cosec\theta-\cot\theta&if\;\pi<\theta<2\pi\end{array}\right.\\(d)\;\left\{\begin{array}{1 1}cosec\theta+\cot\theta&if\;0<\theta>\pi\\cosec\theta-\cot\theta&if\;\pi<\theta<2\pi\end{array}\right.\end{array}$

Answer 1

We have $\sqrt{\large\frac{1+\cos\theta}{1-\cos^2\theta}}$

$\Rightarrow \sqrt{\large\frac{(1+\cos\theta)^2}{1-\cos^2\theta}}$

$\Rightarrow \large\frac{1+\cos\theta}{\sqrt{\sin^2\theta}}$

$\large\frac{1+\cos\theta}{\mid \sin\theta\mid}=\left\{\begin{array}{1 1}\large\frac{1+\cos\theta}{\sin\theta}&\normalsize if\;0\;<\;\theta\;<\pi\\\large\frac{1+\cos\theta}{-\sin\theta}&\normalsize if\;\pi\;<\;\theta\;<2\pi\end{array}\right.$

$\Rightarrow \left\{\begin{array}{1 1}cosec\theta+\cot\theta&if\;0<\theta>\pi\\-cosec\theta-\cot\theta&if\;\pi<\theta<2\pi\end{array}\right.$

Hence (c) is the correct answer.