(2cos^2A-1)^2÷cos^4A-sin^4=1-2sin^2A(2 \cos(2) a – 1) {}^{2} \div \cos(4) – \sin(4) = 1 – 2 \sin(2) a(2cos(2)a−1)2÷cos(4)−sin(4)=1−2sin(2)a Okie le >•< About the author Amaya
Answer: [tex][/tex] Force is push or pull acting on a body which tends to change its state of rest or of motion. It is denoted by “F”. Step-by-step explanation: hope it’s helpful Reply
[tex]{\huge{\boxed{\sf{\blue{❥✰Correct Question✰}}}}}[/tex] [tex]\longmapsto \: \sf\dfrac{{(2cos^2A – 1)}^{2}}{cos^4A – sin^4A} =\: 1 – 2sin^2A⟼ cos 4 A−sin 4 A(2cos 2 A−1) 2 =1−2sin 2 A[/tex] [tex]{\huge{\underline{\bf{\red{❥✰➵ANSWER :-✰}}}}}[/tex] Taking LHS : [tex]\dashrightarrow \sf \dfrac{{(2cos^2A – 1)}^{2}}{cos^4A – sin^4A}⇢ cos 4 A−sin 4 A(2cos 2 A−1) 2[/tex] [tex]\implies \sf \dfrac{{(2cos^2A – 1)}^{2}}{{(cos^2A)}^{2} – {(sin^2A)}^{2}}⟹ (cos 2 A) 2 −(sin 2 A) 2 (2cos 2 A−1) 2[/tex] [tex]\begin{gathered}\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\\\end{gathered}[/tex] [tex]⟹ (cos 2 A+sin 2 A)(cos 2 A−sin 2 A)(2cos 2 A−1) 2 a 2 −b 2 =(a+b)(a−b[/tex] [tex]\implies \sf \dfrac{{(2cos^2A – 1)}^{2}}{(cos^2A – sin^2A)}⟹ (cos 2 A−sin 2 A)(2cos 2 A−1) 2[/tex] [tex]\implies \sf \dfrac{(2 – 2sin^2A – 1)^2}{1 – 2sin^2A}⟹ 1−2sin 2 A(2−2sin 2 A−1) 2[/tex] [tex]\implies \sf \dfrac{(2 – 1 – 2sin^2A)^2}{1 – 2sin^2A}⟹ 1−2sin 2 A(2−1−2sin 2 A) 2[/tex] [tex]\implies \sf \dfrac{(1 – 2sin^2A)^2}{1 – 2sin^2A}⟹ 1−2sin 2 A(1−2sin 2 A) 2 \implies \sf \dfrac{\cancel{(1 – 2sin^2A)}(1 – 2sin^2A)}{\cancel{1 – 2sin^2A}}⟹ 1−2sin 2 A (1−2sin 2 A) (1−2sin 2 A)[/tex] [tex]\implies \sf\bold{\red{1 – 2sin^2A}} \: \: \bigg\lgroup \bold{LHS}\bigg \rgroup⟹1−2sin[/tex] Again, taking RHS : [tex]\dashrightarrow \sf 1 – 2sin^2A⇢1−2sin 2 A[/tex] [tex]\implies\sf\bold{\red{1 – 2sin^2A}} \: \: \bigg\lgroup \bold{RHS}\bigg \rgroup⟹1−2sin 2 A[/tex] [tex]{\large{\pink{\bold{\underline{\leadsto\: LHS =\: RHS}}}}} ⇝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\: :-}}♢TrigonometryIdentities:−\sf cos^2\theta + sin^2\theta =\: 1cos 2 θ+sin 2 θ=1\sf 1 + tan^2\theta =\: sec^2\theta1+tan 2 θ=sec 2 θ\sf 1 + cot^2\theta =\: cosec^2\theta1+cot 2 θ=cosec 2 θ[/tex] [tex]\diamondsuit \: \sf\bold{\purple{Trigonometry \: Complementary\: Angle\: Identities\: :-}}♢[/tex] [tex]\sf sin(90 – \theta) =\: cos\thetasin(90−θ)=cosθ\sf cos(90 – \theta) =\: sin\thetacos(90−θ)=sinθ\sf tan(90 – \theta) =\: cot\thetatan(90−θ)=cotθ\sf cot(90 – \theta) =\: tan\thetacot(90−θ)=tanθ\sf sec(90 – \theta) =\: cosec\thetasec(90−θ)=cosecθ\sf cosec(90 – \theta) =\: sec\thetacosec(90−θ)=secθ[/tex] Thanks! TheEmeraldBoyy Reply
Answer:
[tex][/tex]
Step-by-step explanation:
hope it’s helpful
[tex]{\huge{\boxed{\sf{\blue{❥✰Correct Question✰}}}}}[/tex]
[tex]\longmapsto \: \sf\dfrac{{(2cos^2A – 1)}^{2}}{cos^4A – sin^4A} =\: 1 – 2sin^2A⟼ cos 4 A−sin 4 A(2cos 2 A−1) 2 =1−2sin 2 A[/tex]
[tex]{\huge{\underline{\bf{\red{❥✰➵ANSWER :-✰}}}}}[/tex]
Taking LHS :
[tex]\dashrightarrow \sf \dfrac{{(2cos^2A – 1)}^{2}}{cos^4A – sin^4A}⇢ cos 4 A−sin 4 A(2cos 2 A−1) 2[/tex]
[tex]\implies \sf \dfrac{{(2cos^2A – 1)}^{2}}{{(cos^2A)}^{2} – {(sin^2A)}^{2}}⟹ (cos 2 A) 2 −(sin 2 A) 2 (2cos 2 A−1) 2[/tex]
[tex]\begin{gathered}\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\\\end{gathered}[/tex]
[tex]⟹ (cos 2 A+sin 2 A)(cos 2 A−sin 2 A)(2cos 2 A−1) 2 a 2 −b 2 =(a+b)(a−b[/tex]
[tex]\implies \sf \dfrac{{(2cos^2A – 1)}^{2}}{(cos^2A – sin^2A)}⟹ (cos 2 A−sin 2 A)(2cos 2 A−1) 2[/tex]
[tex]\implies \sf \dfrac{(2 – 2sin^2A – 1)^2}{1 – 2sin^2A}⟹ 1−2sin 2 A(2−2sin 2 A−1) 2[/tex]
[tex]\implies \sf \dfrac{(2 – 1 – 2sin^2A)^2}{1 – 2sin^2A}⟹ 1−2sin 2 A(2−1−2sin 2 A) 2[/tex]
[tex]\implies \sf \dfrac{(1 – 2sin^2A)^2}{1 – 2sin^2A}⟹ 1−2sin 2 A(1−2sin 2 A) 2 \implies \sf \dfrac{\cancel{(1 – 2sin^2A)}(1 – 2sin^2A)}{\cancel{1 – 2sin^2A}}⟹ 1−2sin 2 A (1−2sin 2 A) (1−2sin 2 A)[/tex]
[tex]\implies \sf\bold{\red{1 – 2sin^2A}} \: \: \bigg\lgroup \bold{LHS}\bigg \rgroup⟹1−2sin[/tex]
Again, taking RHS :
[tex]\dashrightarrow \sf 1 – 2sin^2A⇢1−2sin 2 A[/tex]
[tex]\implies\sf\bold{\red{1 – 2sin^2A}} \: \: \bigg\lgroup \bold{RHS}\bigg \rgroup⟹1−2sin 2 A[/tex]
[tex]{\large{\pink{\bold{\underline{\leadsto\: LHS =\: RHS}}}}} ⇝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\: :-}}♢TrigonometryIdentities:−\sf cos^2\theta + sin^2\theta =\: 1cos 2 θ+sin 2 θ=1\sf 1 + tan^2\theta =\: sec^2\theta1+tan 2 θ=sec 2 θ\sf 1 + cot^2\theta =\: cosec^2\theta1+cot 2 θ=cosec 2 θ[/tex]
[tex]\diamondsuit \: \sf\bold{\purple{Trigonometry \: Complementary\: Angle\: Identities\: :-}}♢[/tex]
[tex]\sf sin(90 – \theta) =\: cos\thetasin(90−θ)=cosθ\sf cos(90 – \theta) =\: sin\thetacos(90−θ)=sinθ\sf tan(90 – \theta) =\: cot\thetatan(90−θ)=cotθ\sf cot(90 – \theta) =\: tan\thetacot(90−θ)=tanθ\sf sec(90 – \theta) =\: cosec\thetasec(90−θ)=cosecθ\sf cosec(90 – \theta) =\: sec\thetacosec(90−θ)=secθ[/tex]
Thanks!
TheEmeraldBoyy