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Mathematics

Prove the following identities :

(1−2 sin2A)2cos4A−sin4A\dfrac{(1 - \text{2 sin}^2 A)^2}{\text{cos}^4 A - \text{sin}^4 A} = 2 cos2 A - 1

Trigonometric Identities

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Answer

Solving L.H.S. of the above equation :

⇒(1−2 sin2A)2cos4A−sin4A⇒(1−2 sin2A)2(cos2A−sin2A)(cos2A+sin2A)\Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{\text{cos}^4 A - \text{sin}^4 A} \\[1em] \Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{(\text{cos}^2 A - \text{sin}^2 A)(\text{cos}^2 A + \text{sin}^2 A)}

By formula,

cos2 A + sin2 A = 1 and cos2 A = 1 - sin2 A.

⇒(1−2 sin2A)2(1−sin2A−sin2A)⇒(1−2 sin2A)2(1−2 sin2A)⇒1 - 2 sin2A⇒1−2(1 - cos2A)⇒1−2+2 cos2A⇒2 cos2A−1.\Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{(1 - \text{sin}^2 A - \text{sin}^2 A)} \\[1em] \Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{(1 - \text{2 sin}^2 A)}\\[1em] \Rightarrow \text{1 - 2 sin}^2 A \\[1em] \Rightarrow 1 - 2(\text{1 - cos}^2 A) \\[1em] \Rightarrow 1 - 2 + \text{2 cos}^2 A \\[1em] \Rightarrow \text{2 cos}^2 A - 1.

Since, L.H.S. = R.H.S.

Hence, proved that (1−2 sin2A)2cos4A−sin4A\dfrac{(1 - \text{2 sin}^2 A)^2}{\text{cos}^4 A - \text{sin}^4 A} = 2 cos2 A - 1.

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