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Prove that $\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{\sec A - \tan A}{\sec A + \tan A}$
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Solution: (a) LHS = $\frac{\frac{\cos A}{\sin A} - \cos A}{\frac{\cos A}{\sin A} + \cos A}$ (1 Mark)
= $\frac{\cos A (\frac{1}{\sin A} - 1)}{\cos A (\frac{1}{\sin A} + 1)}$
= $\frac{1 - \sin A}{1 + \sin A}$ (1/2 Mark)
= $\frac{\frac{1}{\cos A} - \frac{\sin A}{\cos A}}{\frac{1}{\cos A} + \frac{\sin A}{\cos A}}$ (1 Mark)
= $\frac{\sec A - \tan A}{\sec A + \tan A}$ = RHS (1/2 Mark)
= $\frac{\cos A (\frac{1}{\sin A} - 1)}{\cos A (\frac{1}{\sin A} + 1)}$
= $\frac{1 - \sin A}{1 + \sin A}$ (1/2 Mark)
= $\frac{\frac{1}{\cos A} - \frac{\sin A}{\cos A}}{\frac{1}{\cos A} + \frac{\sin A}{\cos A}}$ (1 Mark)
= $\frac{\sec A - \tan A}{\sec A + \tan A}$ = RHS (1/2 Mark)