Question

Prove that tan^2A-sin^2A=tan^2A*sin^2A

Answer 1

Answer:

taking LHS and then prove

Answer 2

$\sf{\underline{\boxed{\purple{\large{\bold{Solution }}}}}}$

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$\sf :\implies\:{\bold{ tan^2A - Sin^2A = tan^2A . Sin^2A}}$

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$\sf{\red{\boxed{\bold{TanA = \dfrac{Sin A}{ Cos A} }}}}$⠀

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

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$\sf :\implies\:{\bold{ \dfrac{Sin^2A - Sin^2ACos^2A}{Cos^2A} = tan^2A . Sin^2A }}$

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

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

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$\sf :\implies\:{\bold{ \dfrac{Sin^2A \times Sin^2A}{Cos^2A} = tan^2A . Sin^2A }}$

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$\sf{\red{\boxed{\bold{\dfrac{Sin^2A}{Cos^2A} = Tan^2A}}}}$

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$\sf :\implies\:{\bold{ Tan^2A . Sin^2A = tan^2A . Sin^2A }}$

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LHS = RHS

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⠀⠀⠀⠀⠀⠀⠀⠀.... Hence Proved

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