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Question
Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Solution
We have to prove (sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
We know that `sec^2 theta A - tan^2 theta A = 1`
So, we have
(sec A + tan A - 1)(sec A - tan A + 1) = {sec A + (tan A - 1)}{sec A - (tan A - 1)}
`= sec^2 A - (tan A - 1)^2`
`= sec^2 A - (tan^2 A - 2 tan A + 1)`
`= (sec^2 A - tan^2 A) + 2 tan A - 1`
So we have
(sec A + tan A - 1)(sec A - tan A + 1) = 1 + tan A - 1
= 2 tan A
Hence proved.
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