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Question

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

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Solution

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

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