A) \[\frac{\cos A}{\sqrt{1-{{\cos }^{2}}A}}\]
B) \[\frac{\sec A}{\sqrt{1-{{\sec }^{2}}A}}\]
C) \[\frac{\sin A}{\sqrt{1-{{\sin }^{2}}A}}\]
D) \[\frac{1}{\sqrt{1-{{\sin }^{2}}A}}\]
Correct Answer: C
Solution :
[c]\[\tan A=\frac{\sin A}{\cos A}=\frac{\sin A}{\sqrt{1-{{\sin }^{2}}A}}\]
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