Question

(sec A+ tan A - 1) (sec A- tan A + 1) = 2 tan A​

Answer 1

Answer:

$LHS = (secA+tanA-1)(secA-tanA+1)\\= [secA+(tanA-1)][ secA-(tanA-1)]\\=sec^{2}A - (tanA-1)^{2}$

$\boxed {\pink { (x+y)(x-y) = x^{2} - y^{2} }}$

$= sec^{2}A - ( tan^{2} A + 1^{2} - 2tanA )\\= sec^{2}A - ( sec^{2} A - 2tanA )$

$\boxed { \pink { tan^{2}A + 1 = sec^{2} A }}$

$= sec^{2} A - sec^{2} A + 2tanA$

$= 2tanA$

$= RHS$

$Hence , proved$

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