A) \[\frac{1}{2}{{\sec }^{2}}\frac{x}{2}\]
B) \[-\frac{1}{2}{{\sec }^{2}}\frac{x}{2}\]
C) \[{{\sec }^{2}}\frac{x}{2}\]
D) \[-{{\sec }^{2}}\frac{x}{2}\]
Correct Answer: A
Solution :
Let \[y=\frac{\sqrt{1-\cos \,x}}{\sqrt{1+\cos \,x}}=\sqrt{\frac{1-1+2{{\sin }^{2}}\frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}-1+1}}=\tan \left( \frac{x}{2} \right)\] \[\therefore \,\,\frac{dy}{dx}={{\sec }^{2}}\frac{x}{2}.\frac{1}{2}=\frac{1}{2}{{\sec }^{2}}\frac{x}{2}\]
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