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What is the value of $$1 + \frac{{{{\tan }^2}A}}{{1 + \sec A}}?$$
Answer & Solution
Correct Answer:
Option
C
$$\eqalign{
& 1 + \frac{{{{\tan }^2}A}}{{1 + \sec A}} \cr
& = 1 + \frac{{{{\sec }^2}A - 1}}{{1 + \sec A}} \cr
& = \frac{{1 + \sec A + {{\sec }^2}A - 1}}{{1 + \sec A}} \cr
& = \frac{{\sec A\left( {1 + \sec A} \right)}}{{1 + \sec A}} \cr
& = \sec A \cr} $$
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