Q109 / 200 · Manipal (MET) 2012 · Mathematics

If $\frac{\pi}{2} \lt \theta \lt \frac{3 \pi}{2}$, then the value of $\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$ is equal to

  • Correct answer
    A
    $-\sec \theta+\tan \theta$
  • B
    $\sec \theta+\tan \theta$
  • C
    $\sec \theta-\tan \theta$
  • D
    None of these

Solution

$\begin{aligned} \sqrt{\frac{1-\sin \theta}{1+\sin \theta}} & =\sqrt{\frac{(1-\sin \theta)^2}{1-\sin ^2 \theta}} \\ & =\frac{1-\sin \theta}{\sqrt{\cos ^2 \theta}}=\frac{1-\sin \theta}{|\cos \theta|} \\ & =\frac{1-\sin \theta}{-\cos \theta} \quad\left(\because \frac{\pi}{2} \lt \theta \lt \frac{3 \pi}{2}\right) \\ & =-\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta} \\ & =-\sec \theta+\tan \theta\end{aligned}$

Hence, the correct answer is option: 'A'.

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