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

$\begin{array}{1 1}(a)\;\left\{\begin{array}{1 1}\sec\theta-\tan\theta,&if\;-\large\frac{\pi}{2}\;< \theta\;<\large\frac{\pi}{2}\\-\sec\theta+\tan\theta&if\;\large\frac{\pi}{2}\; <\theta<\large\frac{3\pi}{2}\end{array}\right.\\(b)\;\left\{\begin{array}{1 1}\sec\theta-\tan\theta,&if\;-\large\frac{3\pi}{2}\;< \theta\;<\large\frac{\pi}{2}\\-\sec\theta+\tan\theta&if\;\large\frac{\pi}{2}\; <\theta<\large\frac{3\pi}{2}\end{array}\right.\\(c)\;\left\{\begin{array}{1 1}\sec\theta-\tan\theta,&if\;\large\frac{\pi}{2}\;< \theta\;<\large\frac{3\pi}{2}\\-\sec\theta+\tan\theta&if\;\large\frac{\pi}{2}\; <\theta<\large\frac{3\pi}{2}\end{array}\right.\\(d)\;None\;of\;these\end{array}$

1 Answer

We have $\sqrt{\large\frac{1-\sin\theta}{1+\sin\theta}}$
$\Rightarrow \sqrt{\large\frac{(1-\sin\theta)^2}{1-\sin^2\theta}}$
$\Rightarrow \large\frac{1-\sin\theta}{\sqrt{\cos^2\theta}}$
$\large\frac{1-\sin\theta}{\mid \cos\theta\mid}=\left\{\begin{array}{1 1}\large\frac{1-\sin\theta}{\cos\theta}&if\;-\large\frac{\pi}{2}\;<\;\theta\;\large\frac{\pi}{2}\\\large\frac{1-\sin\theta}{-\cos\theta}&if\;\large\frac{\pi}{2}\;<\theta\;<\large\frac{3\pi}{2}\end{array}\right.$
$\left\{\begin{array}{1 1}\sec\theta-\tan\theta,&if\;-\large\frac{\pi}{2}\;< \theta\;<\large\frac{\pi}{2}\\-\sec\theta+\tan\theta&if\;\large\frac{\pi}{2}\; <\theta<\large\frac{3\pi}{2}\end{array}\right.$
Hence (a) is the correct answer.
answered Oct 14, 2013 by sreemathi.v
 

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