Question
(sec A+tan A−1)(sec A−tan A+1)tan A=
Answer: Option D
:
D
(sec A+tan A−1)(sec A−tan A+1)tan A
=[sec A+(tan A−1)][sec A−(tan A−1)]tan A
=sec2A−(tan A−1)2tan A
=sec2A−(tan2A−2tan A+1)tan A
=sec2A−tan2A+2tan A−1tan A
=1+2tan A−1tan A……(∵1+tan2A=sec2A)
=2tan Atan A
=2
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:
D
(sec A+tan A−1)(sec A−tan A+1)tan A
=[sec A+(tan A−1)][sec A−(tan A−1)]tan A
=sec2A−(tan A−1)2tan A
=sec2A−(tan2A−2tan A+1)tan A
=sec2A−tan2A+2tan A−1tan A
=1+2tan A−1tan A……(∵1+tan2A=sec2A)
=2tan Atan A
=2
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