Question

Prove the following: sec^(4)A-1=2tan^(2)A+tan^(4)A

Answer 1

$\large \green{ \underline{question}} \\ \\ \large \implies{ \sec {}^{4} ( \: A) - 1 = 2 \: \tan {}^{2} ( \:A ) + \: \tan {}^{4} ( \: A) } \\ \\ \large \implies{ \underline{to \: prove \: \: LHS \: = \: RHS \: }} \\ \\ \large \implies{ \underline{from \: \: LHS \: \: and \: \: RHS}} \\ \\ \large \implies{ \sec {}^{4} ( \:A ) - \tan {}^{4} ( \:A ) = 1 \: + \: 2 \tan {}^{2} ( \:A ) } \\ \\ \large \implies{ \underline{LHS}} \\ \\ \large \implies{ \sec {}^{4} ( \:A) - \tan {}^{4} ( \: A) } \\ \\ \large \implies{( \sec {}^{2} ( \:A) ) {}^{2} - ( \tan {}^{2} ( \:A) ) {}^{2} } \\ \\ \large \implies{it \: \: is \: \: in \: \: form \: of } \\ \\ \large \implies{{a}^{2} - {b}^{2} = (a + b)(a - b) } \\ \\ \large \implies{( \sec {}^{2} ( \:A ) + \: \tan {}^{2} ( \:A) ) \: \: ( \sec {}^{2} ( \:A ) - \: \tan {}^{2} ( \: A) {}^{2} } \\ \\ \large \implies{ \sec {}^{2} ( \: A) + \tan {}^{2} ( \: A ) \times \: 1} \\ \\ \large \implies{ \tan {}^{2} ( \:A ) \: + \: 1 + \: \tan {}^{2} ( \:A) } \\ \\ \large \implies{ \boxed{1 \: + \: 2 \tan {}^{2} ( \: A) = \: RHS}}$

Formula used

1)=> ( a - b) ( a + b) = a^2 - b^2

2)=> sec^2A - tan^2A = 1