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

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

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## 1 Answer

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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.
answered Oct 14, 2013

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1 answer

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1 answer