Discussion
The value of \(\frac{\tan A}{\sec A + 1} + \frac{\sec A + 1}{\tan A}\) is:
Answer: C
Find a common denominator: \(\tan A (\sec A + 1)\).
Expression = \(\frac{\tan^2 A + (\sec A + 1)^2}{\tan A (\sec A + 1)}\).
Numerator = \(\tan^2 A + \sec^2 A + 2\sec A + 1\). Using \(\tan^2 A = \sec^2 A - 1\), it becomes \((\sec^2 A - 1) + \sec^2 A + 2\sec A + 1 = 2\sec^2 A + 2\sec A = 2\sec A(\sec A + 1)\).
So, Expression = \(\frac{2\sec A(\sec A + 1)}{\tan A (\sec A + 1)} = \frac{2\sec A}{\tan A} = \frac{2/\cos A}{\sin A/\cos A} = \frac{2}{\sin A} = 2\csc A\).
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