Solve this

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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