A) \[\frac{1}{2}\sqrt{1+\sec x}\]
B) \[\sqrt{1+\sec x}\]
C) \[-\frac{1}{2}\sqrt{1+\sec x}\]
D) \[-\sqrt{1+\sec x}\]
Correct Answer: A
Solution :
\[\frac{d}{dx}{{\cos }^{-1}}\sqrt{\cos x}=\frac{\sin x}{2\sqrt{\cos x}\sqrt{1-\cos x}}\] \[=\frac{\sqrt{1-{{\cos }^{2}}x}}{2\sqrt{\cos x}\sqrt{1-\cos x}}=\frac{1}{2}\sqrt{\frac{1+\cos x}{\cos x}}\].
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