(C) Let the expression be $E = (\sec A + \tan A - 1)(\sec A - (\tan A - 1)) - 2\tan A$. Using the identity $(x+y)(x-y) = x^2 - y^2$,we rewrite the fir

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(C) Let the expression be $E = (\sec A + \tan A - 1)(\sec A - (\tan A - 1)) - 2\tan A$. Using the identity $(x+y)(x-y) = x^2 - y^2$,we rewrite the first part: $E = (\sec A + (\tan A - 1))(\sec A - (\tan A - 1)) - 2\tan A$ $E = \sec^2 A - (\tan A - 1)^2 - 2\tan A$ $E = \sec^2 A - (\tan^2 A - 2\tan A + 1) - 2\tan A$ $E = \sec^2 A - \tan^2 A + 2\tan A - 1 - 2\tan A$ Since $\sec^2 A - \tan^2 A = 1$,we have: $E = 1 - 1 = 0$.

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Mix Examples-Trigonometrical Ratios, Functions and Identities

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

A
$\sec A$
B
$2\sec A$
C
$0$
D
$1$

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Trigonometrical Ratios, Functions and Identities

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Mix Examples-Trigonometrical Ratios, Functions and Identities

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

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