A) \[\frac{\sec A}{1-\cos A}\]
B) \[\frac{\sec A}{1+{{\cos }^{2}}A}\]
C) \[\frac{\sec A}{1+\cos A}\]
D) None of these
Correct Answer: C
Solution :
| \[\frac{\tan A-\sin A}{{{\sin }^{3}}A}=\frac{\frac{\sin A}{\cos A}-\sin A}{{{\sin }^{3}}A}\] |
| \[=\frac{(1-\cos A)}{\cos A\cdot {{\sin }^{2}}A}\times \frac{(1+\cos A)}{1+(\cos A)}\] |
| \[=\frac{(1-{{\cos }^{2}}A)}{\cos A\cdot {{\sin }^{2}}A(1+\cos A)}\] |
| \[=\frac{{{\sin }^{2}}A}{\cos A\cdot {{\sin }^{2}}A(1+\cos A)}\] |
| \[=\frac{1}{\cos A}\cdot \frac{1}{1+\cos A}=\frac{\sec A}{1+\cos A}\] |
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