A) \[\frac{-1}{2\left| x \right|\sqrt{{{x}^{2}}-1}}\]
B) \[\frac{-1}{2x\sqrt{{{x}^{2}}-1}}\]
C) \[\frac{1}{2x\sqrt{{{x}^{2}}-1}}\]
D) none of these
Correct Answer: A
Solution :
[a] Let \[x=\sec \theta .\]Then \[y={{\tan }^{-1}}\sqrt{\frac{\sec \theta +1}{\sec \theta -1}}\] \[={{\tan }^{-1}}\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}={{\tan }^{-1}}\left( \cot \frac{\theta }{2} \right)\] \[={{\tan }^{-1}}\left\{ \tan \left( \frac{\pi }{2}-\frac{\theta }{2} \right) \right\}\] \[=\frac{\pi }{2}-\frac{1}{2}{{\sec }^{-1}}x\] \[\therefore \frac{dy}{dx}=-\frac{1}{2}.\frac{1}{\left| x \right|\sqrt{{{x}^{2}}-1}}\]
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