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Question

# If $\mathrm{cot}x\left(1+\mathrm{sin}x\right)=4m\mathrm{and}\mathrm{cot}x\left(1-\mathrm{sin}x\right)=4n,$ prove that ${\left({m}^{2}+{n}^{2}\right)}^{2}=mn$.

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Solution

## $\mathrm{Given}:\phantom{\rule{0ex}{0ex}}4m=cotx\left(1+\mathrm{sin}x\right)and4n=cotx\left(1-\mathrm{sin}x\right)\phantom{\rule{0ex}{0ex}}\mathrm{Multiplying}\mathrm{both}\mathrm{the}\mathrm{equations}:\phantom{\rule{0ex}{0ex}}⇒16mn=co{t}^{2}x\left(1-{\mathrm{sin}}^{2}x\right)\phantom{\rule{0ex}{0ex}}⇒16mn=co{t}^{2}x.{\mathrm{cos}}^{2}x\phantom{\rule{0ex}{0ex}}⇒mn=\frac{{\mathrm{cos}}^{4}x}{16{\mathrm{sin}}^{2}x}\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{the}\mathrm{given}\mathrm{equation}:\phantom{\rule{0ex}{0ex}}16{m}^{2}=co{t}^{2}x{\left(1+\mathrm{sin}x\right)}^{2}and16{n}^{2}=co{t}^{2}x{\left(1-\mathrm{sin}x\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒16{m}^{2}-16{n}^{2}=co{t}^{2}x\left(4\mathrm{sin}x\right)\phantom{\rule{0ex}{0ex}}⇒{m}^{2}-{n}^{2}=\frac{co{t}^{2}x.\mathrm{sin}x}{4}\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{both}\mathrm{sides},\phantom{\rule{0ex}{0ex}}{\left({m}^{2}-{n}^{2}\right)}^{2}=\frac{co{t}^{4}x.{\mathrm{sin}}^{2}x}{16}\phantom{\rule{0ex}{0ex}}⇒{\left({m}^{2}-{n}^{2}\right)}^{2}=\frac{{\mathrm{cos}}^{4}x}{16{\mathrm{sin}}^{2}x}\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(1\right)\mathrm{and}\left(2\right):\phantom{\rule{0ex}{0ex}}{\left({m}^{2}-{n}^{2}\right)}^{2}=\mathit{}mn\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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