(sec A + tan A - 1)(sec A - tan A + 1) - 2tan | vedclass

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Question

(A) Given expression: $(\sec A + 	an A - 1)(\sec A - 	an A + 1) - 2	an A$Rewrite the first part as: $(\sec A + (	an A - 1))(\sec A - (	an A - 1))

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(A) Given expression: $(\sec A + 	an A - 1)(\sec A - 	an A + 1) - 2	an A$Rewrite the first part as: $(\sec A + (	an A - 1))(\sec A - (	an A - 1)) - 2	an A$Using the identity $(x+y)(x-y) = x^2 - y^2$,where $x = \sec A$ and $y = 	an A - 1$:$= \sec^2 A - (	an A - 1)^2 - 2	an A$$= \sec^2 A - (	an^2 A - 2	an A + 1) - 2	an A$$= \sec^2 A - 	an^2 A + 2	an A - 1 - 2	an A$Since $\sec^2 A - 	an^2 A = 1$:$= 1 - 1 = 0$

$(\sec A + \tan A - 1)(\sec A - \tan A + 1) - 2\tan A = $

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