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Answer:
[tex]\longmapsto \: \sf\dfrac{{(2cos^2A - 1)}^{2}}{cos^4A - sin^4A} =\: 1 - 2sin^2A[/tex]
Taking LHS :
[tex]\dashrightarrow \sf \dfrac{{(2cos^2A - 1)}^{2}}{cos^4A - sin^4A}[/tex]
[tex]\implies \sf \dfrac{{(2cos^2A - 1)}^{2}}{{(cos^2A)}^{2} - {(sin^2A)}^{2}}[/tex]
[tex]\implies \sf \dfrac{{(2cos^2A - 1)}^{2}}{(cos^2A + sin^2A)(cos^2A - sin^2A)} \: \bigg\lgroup a^2 - b^2 =\: (a + b)(a - b)\bigg \rgroup\\[/tex]
[tex]\implies \sf \dfrac{{(2cos^2A - 1)}^{2}}{(cos^2A - sin^2A)}[/tex]
[tex]\implies \sf \dfrac{\{2(1 - sin^2A)- 1\}^2}{1 - sin^2A - sin^2A} \: \bigg\lgroup cos^2A =\: 1 - sin^2A\bigg \rgroup\\[/tex]
[tex]\implies \sf \dfrac{(2 - 2sin^2A - 1)^2}{1 - 2sin^2A}[/tex]
[tex]\implies \sf \dfrac{(2 - 1 - 2sin^2A)^2}{1 - 2sin^2A}[/tex]
[tex]\implies \sf \dfrac{(1 - 2sin^2A)^2}{1 - 2sin^2A}[/tex]
[tex]\implies \sf \dfrac{\cancel{(1 - 2sin^2A)}(1 - 2sin^2A)}{\cancel{1 - 2sin^2A}}[/tex]
[tex]\implies \sf\bold{\red{1 - 2sin^2A}} \: \: \bigg\lgroup \bold{LHS}\bigg \rgroup[/tex]
Again, taking RHS :
[tex]\dashrightarrow \sf 1 - 2sin^2A[/tex]
[tex]\implies\sf\bold{\red{1 - 2sin^2A}} \: \: \bigg\lgroup \bold{RHS}\bigg \rgroup[/tex]
[tex]{\large{\pink{\bold{\underline{\leadsto\: LHS =\: RHS}}}}}[/tex]
[tex]\clubsuit \: \sf\boxed{\bold{\green{Hence, Proved}}}[/tex]
[tex]\rule{150}{2}[/tex]
Extra Formula Related to Trigonometry :
[tex]\diamondsuit\: \sf\bold{\purple{Trigonometry\: Identities\: :-}}[/tex]
[tex]\sf cos^2\theta + sin^2\theta =\: 1[/tex]
[tex]\sf 1 + tan^2\theta =\: sec^2\theta[/tex]
[tex]\sf 1 + cot^2\theta =\: cosec^2\theta[/tex]
[tex]\diamondsuit \: \sf\bold{\purple{Trigonometry \: Complementary\: Angle\: Identities\: :-}}[/tex]
[tex]\sf sin(90 - \theta) =\: cos\theta[/tex]
[tex]\sf cos(90 - \theta) =\: sin\theta[/tex]
[tex]\sf tan(90 - \theta) =\: cot\theta[/tex]
[tex]\sf cot(90 - \theta) =\: tan\theta[/tex]
[tex]\sf sec(90 - \theta) =\: cosec\theta[/tex]
[tex]\sf cosec(90 - \theta) =\: sec\theta[/tex]