A) \[\sec A\]
B) \[2\sec A\]
C) 0
D) 1
Correct Answer: C
Solution :
\[(\sec A+\tan A-1)(\sec A-\tan A+1)-2\tan A\] \[=({{\sec }^{2}}A-{{\tan }^{2}}A)+\sec A+\tan A-\sec A\] \[+\tan A-1-2\tan A=0\] \[(\because {{\sec }^{2}}A-{{\tan }^{2}}A=1)\]
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