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Mathematics

Prove the following identities :

tan2 A - tan2 B = sin2A−sin2Bcos2A.cos2B\dfrac{\text{sin}^2 A - \text{sin}^2 B}{\text{cos}^2 A. \text{cos}^2 B}

Trigonometric Identities

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Answer

Solving L.H.S. of the equation :

⇒sin2Acos2A−sin2Bcos2B⇒sin2A. cos2B−sin2B. cos2Acos2A. cos2B⇒sin2A(1− sin2B)−sin2B(1−sin2A)cos2A. cos2B⇒sin2A−sin2A. sin2B−sin2B+ sin2A. sin2Bcos2A. cos2B⇒sin2A− sin2Bcos2A. cos2B\Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} - \dfrac{\text{sin}^2 B}{\text{cos}^2 B} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A. \text{ cos}^2 B - \text{sin}^2 B. \text{ cos}^2 A}{\text{cos}^2 A. \text{ cos}^2 B} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A(1 - \text{ sin}^2 B) - \text{sin}^2 B(1 - \text{sin}^2 A)}{\text{cos}^2 A. \text{ cos}^2 B} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A - \text{sin}^2 A. \text{ sin}^2 B - \text{sin}^2 B + \text{ sin}^2 A. \text{ sin}^2 B}{\text{cos}^2 A. \text{ cos}^2 B} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A - \text{ sin}^2 B}{\text{cos}^2 A. \text{ cos}^2 B}

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

Hence, proved that tan2 A - tan2 B = sin2A−sin2Bcos2A.cos2B\dfrac{\text{sin}^2 A - \text{sin}^2 B}{\text{cos}^2 A. \text{cos}^2 B}.

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