Question

(2cos^2A-1)^2÷cos^4A-sin^4=1-2sin^2A
$(2 \cos(2) a - 1) {}^{2} \div \cos(4) - \sin(4) = 1 - 2 \sin(2) a$
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Answer 1

Answer:-

We have to prove:-

$\implies \sf \:\dfrac{(2 { \cos }^{2} A- 1) ^{2} }{ { \cos }^{4} A - \sin ^{4} A } = 1 - 2 { \sin }^{2} A \\ \\ \\ \implies \sf \: \frac{(2 { \cos }^{2} A - 1) ^{2}}{ { ({ \cos}^{2} A) }^{2} - { ({ \sin}^{2} A)}^{2} } = 1- 2 { \sin }^{2} A$

Using a² - b² = (a + b)(a - b) we get,

$\implies \sf \: \dfrac{ { {(2 \cos }^{2} A - 1)}^{2} }{( { \cos }^{2} A + { \sin}^{2} A)({ \cos }^{2} A - { \sin}^{2} A)} = 1 - 2 { \sin}^{2} A$

Using the identity sin² A + cos² A = 1 we get,

$\implies \sf \: \dfrac{(2 { \cos}^{2} A - 1) ^{2} }{ { \cos }^{2} A - { \sin }^{2} A } = 1 - 2 { \sin }^{2} A$

Using cos² A = 1 - sin² A in LHS we get,

$\implies \sf \: \dfrac{( {2 (1 - \sin}^{2} A) - 1) ^{2} }{1 - { \sin }^{2} A - { \sin}^{2} A} = 1 - 2 { \sin }^{2} A \\ \\ \\ \implies \sf \: \frac{ {(2 - 2 { \sin }^{2} A - 1) }^{2} }{(1 - 2 { \sin }^{2} A)} = 1 - 2 { \sin }^{2} A \\ \\ \\ \implies \sf \: \frac{(1 - 2 { \sin }^{2} A) ^{2} }{(1 - 2 { \sin }^{2} A) } = 1 - 2 { \sin }^{2} A \\ \\ \\ \implies \sf \: \frac{(1 - 2 { \sin }^{2} A)(1 - 2 { \sin }^{2} A) }{(1 - 2 { \sin }^{2} A) } = 1 - 2 { \sin }^{2} A \\ \\ \\ \implies \sf \: 1 - 2 { \sin }^{2} A = 1 - 2 { \sin }^{2} A$

Hence, Proved.

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Simple Method:-

• cos 2A = 2 cos² A - 1

• cos 2A = cos² A - sin² A = cos⁴ A - sin⁴ A

• cos 2A = 1 - 2sin² A.

So,

⟹ (cos 2A)² / cos 2A = cos 2A

⟹ cos 2A = cos 2A

Hence, Proved.

Answer 2

Answer:

Correct Question :-

$\longmapsto \: \sf\dfrac{{(2cos^2A - 1)}^{2}}{cos^4A - sin^4A} =\: 1 - 2sin^2A$

Solution :-

Taking LHS :

$\dashrightarrow \sf \dfrac{{(2cos^2A - 1)}^{2}}{cos^4A - sin^4A}$

$\implies \sf \dfrac{{(2cos^2A - 1)}^{2}}{{(cos^2A)}^{2} - {(sin^2A)}^{2}}$

$\implies \sf \dfrac{{(2cos^2A - 1)}^{2}}{(cos^2A + sin^2A)(cos^2A - sin^2A)} \: \bigg\lgroup a^2 - b^2 =\: (a + b)(a - b)\bigg \rgroup$

$\implies \sf \dfrac{{(2cos^2A - 1)}^{2}}{(cos^2A - sin^2A)}$

$\implies \sf \dfrac{\{2(1 - sin^2A)- 1\}^2}{1 - sin^2A - sin^2A} \: \bigg\lgroup cos^2A =\: 1 - sin^2A\bigg \rgroup$

$\implies \sf \dfrac{(2 - 2sin^2A - 1)^2}{1 - 2sin^2A}$

$\implies \sf \dfrac{(2 - 1 - 2sin^2A)^2}{1 - 2sin^2A}$

$\implies \sf \dfrac{(1 - 2sin^2A)^2}{1 - 2sin^2A}$

$\implies \sf \dfrac{\cancel{(1 - 2sin^2A)}(1 - 2sin^2A)}{\cancel{1 - 2sin^2A}}$

$\implies \sf\bold{\red{1 - 2sin^2A}} \: \: \bigg\lgroup \bold{LHS}\bigg \rgroup$

Again, taking RHS :

$\dashrightarrow \sf 1 - 2sin^2A$

$\implies\sf\bold{\red{1 - 2sin^2A}} \: \: \bigg\lgroup \bold{RHS}\bigg \rgroup$

${\large{\pink{\bold{\underline{\leadsto\: LHS =\: RHS}}}}}$

$\clubsuit \: \sf\boxed{\bold{\green{Hence, Proved}}}$

$\rule{150}{2}$

Extra Formula Related to Trigonometry :

$\diamondsuit\: \sf\bold{\purple{Trigonometry\: Identities\: :-}}$

• $\sf cos^2\theta + sin^2\theta =\: 1$
• $\sf 1 + tan^2\theta =\: sec^2\theta$
• $\sf 1 + cot^2\theta =\: cosec^2\theta$

$\diamondsuit \: \sf\bold{\purple{Trigonometry \: Complementary\: Angle\: Identities\: :-}}$

• $\sf sin(90 - \theta) =\: cos\theta$
• $\sf cos(90 - \theta) =\: sin\theta$
• $\sf tan(90 - \theta) =\: cot\theta$
• $\sf cot(90 - \theta) =\: tan\theta$
• $\sf sec(90 - \theta) =\: cosec\theta$
• $\sf cosec(90 - \theta) =\: sec\theta$