What is the value of $$1 + \frac{{{{\tan }^2}A}}{{1 + \sec A}}?$$
A. cosecA
B. cosA
C. secA
D. sinA
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & 1 + \frac{{{{\tan }^2}A}}{{1 + \sec A}} \cr & = 1 + \frac{{{{\sec }^2}A - 1}}{{1 + \sec A}} \cr & = \frac{{1 + \sec A + {{\sec }^2}A - 1}}{{1 + \sec A}} \cr & = \frac{{\sec A\left( {1 + \sec A} \right)}}{{1 + \sec A}} \cr & = \sec A \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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