Question:
$\frac{\tan A+\sin A}{\tan A-\sin A}=\frac{\sec A+1}{\sec A-1}$
Solution:
$\frac{\tan A+\sin A}{\tan A-\sin A}$
$=\frac{\frac{\sin A}{\cos A}+\sin A}{\frac{\sin A}{\cos A}-\sin A}$
$=\frac{\sin A\left(\frac{1}{\cos A}+1\right)}{\sin A\left(\frac{1}{\cos A}-1\right)}$
$=\frac{\sec A+1}{\sec A-1} \quad\left(\sec A=\frac{1}{\cos A}\right)$
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