Question

Prove the following trigonometric identities. (sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A

Answer 1

Answer with Step-by-step explanation:

Given :   (sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A

LHS : (sec A + tan A − 1) (sec A − tan A + 1)

= {sec A + (tan A − 1)} {sec A − (tan A - 1)}

= sec²A - (tan A - 1)²

[By using identity , (a + b) (a - b) = a² - b²]

= sec²A - (tan²A - 2tanA + 1²)

[By using identity , (a - b)² = a² - 2ab + b²]

= (sec²A - tan²A) + 2tanA - 1

= 1 + 2tanA - 1

[By using the identity , sec² θ -  tan² θ = 1]

= 2tanA  

(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A

L.H.S = R.H.S  

Hence Proved..

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Answer 2

Answer:

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