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prove tan^2(x)-sin^2(x)=tan^2(x)sin^2(x)

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∂²∂ y∂ x ((xy³−x³y)x²+y² )

Solution

−x⁶−9x⁴y²+9x²y⁴+y⁶(x²+y²)³

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∂²∂ y∂ x ((xy³−x³y)x²+y² )

∂ ∂ y (xy³−x³yx²+y² )=−x⁵+4x³y²+xy⁴(x²+y²)²

=∂ ∂ x (−x⁵+4x³y²+xy⁴(x²+y²)² )

∂ ∂ x (−x⁵+4x³y²+xy⁴(x²+y²)² )=−x⁶−9x⁴y²+9x²y⁴+y⁶(x²+y²)³

=−x⁶−9x⁴y²+9x²y⁴+y⁶(x²+y²)³

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+▭▭	>	4	5	6	×	☐☐☐
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prove tan^2(x)-sin^2(x)=tan^2(x)sin^2(x)

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