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Question

# Prove the following identities (1-16) $\frac{{\mathrm{tan}}^{3}x}{1+{\mathrm{tan}}^{2}x}+\frac{{\mathrm{cot}}^{3}x}{1+{\mathrm{cot}}^{2}x}=\frac{1-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}{\mathrm{sin}x\mathrm{cos}x}$

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Solution

## $\mathrm{LHS}=\frac{{\mathrm{tan}}^{3}x}{1+{\mathrm{tan}}^{2}x}+\frac{{\mathrm{cot}}^{3}x}{1+{\mathrm{cot}}^{2}x}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{tan}}^{3}x}{{\mathrm{sec}}^{2}x}+\frac{{\mathrm{cot}}^{3}x}{{\mathrm{cosec}}^{2}x}\phantom{\rule{0ex}{0ex}}=\frac{\frac{{\mathrm{sin}}^{3}x}{{\mathrm{cos}}^{3}x}}{\frac{1}{{\mathrm{cos}}^{2}x}}+\frac{\frac{{\mathrm{cos}}^{3}x}{{\mathrm{sin}}^{3}x}}{\frac{1}{{\mathrm{sin}}^{2}x}}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{3}x}{{\mathrm{cos}}^{3}x}×\frac{{\mathrm{cos}}^{2}x}{1}+\frac{{\mathrm{cos}}^{3}x}{{\mathrm{sin}}^{3}x}×\frac{{\mathrm{sin}}^{2}x}{1}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{3}x}{\mathrm{cos}x}+\frac{{\mathrm{cos}}^{3}x}{\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}{\mathrm{sin}x\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}=\frac{{\left({\mathrm{sin}}^{2}x\right)}^{2}+{\left({\mathrm{cos}}^{2}x\right)}^{2}}{\mathrm{sin}x\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}=\frac{{\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}\mathrm{x}\right)}^{2}-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}{\mathrm{sin}x\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}=\frac{{1}^{2}-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}{\mathrm{sin}x\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}=\frac{1-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}{\mathrm{sin}x\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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