Question

prove the following​

Answer 1

$\sf{\orange{\underline{\huge{\mathsf{Solution}}}}}$

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$\sf: \implies\: {\bold{ cot^2A - cot^2B = \dfrac{cos^2A - cos^2B}{sin^2A . Sin^2B}}}$

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$\sf{\red{\boxed{\bold{Cot^2\theta = \dfrac{Cos^2\theta}{Sin^2\theta}}}}}$

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$\sf: \implies\: {\bold{ \dfrac{Cos^2A}{Sin^2A} - \dfrac{Cos^2B}{Sin^2B} = \dfrac{cos^2A - cos^2B}{sin^2A . Sin^2B}}}$

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$\sf: \implies\: {\bold{\dfrac{Cos^2A.Sin^2B - Cos^2BSin^2A}{Sin^2B.Sin^2A} = \dfrac{cos^2A - cos^2B}{sin^2A . Sin^2B}}}$

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$\sf{\red{\boxed{\bold{Sin^2\theta = 1 - Cos^2\theta}}}}$

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⠀$\sf: \implies\: {\bold{\dfrac{Cos^2A( 1 - Cos^2B) - Cos^2B( 1 - Cos^2A)}{Sin^2B.Sin^2A} = \dfrac{cos^2A - cos^2B}{sin^2A . Sin^2B}}}$

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$\sf: \implies\: {\bold{\dfrac{Cos^2A - Cos^2A . Cos^2B - Cos^2B + Cos^2B. Cos^2A}{Sin^2B.Sin^2A} = \dfrac{cos^2A - cos^2B}{sin^2A . Sin^2B}}}$

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$\sf: \implies\: {\bold{\dfrac{Cos^2A - \cancel{Cos^2A . Cos^2B} - Cos^2B + \cancel{ Cos^2A . Cos^2B}}{Sin^2B.Sin^2A} = \dfrac{cos^2A - cos^2B}{sin^2A . Sin^2B}}}$

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$\sf: \implies\: {\bold{\dfrac{Cos^2A - Cos^2B }{Sin^2A.Sin^2B} = \dfrac{cos^2A - cos^2B}{sin^2A . Sin^2B}}}$

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀.......Hence Proved

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