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Question

(sec A+tan A−1)(sec A−tan A+1)tan A=


A
\N
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B
tan A
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C
1
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D
2
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Solution

The correct option is D 2

(sec A+tan A1)(sec Atan A+1)tan A

=[sec A+(tan A1)] [sec A(tan A1)]tan A

=sec2 A(tan A1)2tan A

=sec2 A(tan2 A2 tan A+1)tan A

=sec2 Atan2 A+2 tan A1tan A

=1+2 tan A1tan A(1+tan2 A=sec2 A)

=2 tan Atan A

=2


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