Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ 4m=cot\theta \left(1+\sin \theta \right)and4n=cot\theta \left(1-\sin \theta \right) \\ \mathrm{Multiplying}\mathrm{both}\mathrm{the}\mathrm{equations}: \\ \Rightarrow 16mn=co{t}^2\theta \left(1-{\sin }^2\theta \right) \\ \Rightarrow 16mn=co{t}^2\theta .{\cos }^2\theta \\ \Rightarrow mn=\frac{{\cos }^4\theta }{16{\sin }^2\theta }\left(1\right) \\ \mathrm{Squaring}\mathrm{the}\mathrm{given}\mathrm{equation}: \\ 16m^2=co{t}^2\theta {\left(1+\sin \theta \right)}^{2}and16n^2=co{t}^2\theta {\left(1-\sin \theta \right)}^2 \\ \Rightarrow 16m^2-16n^2=co{t}^2\theta \left(4\sin \theta \right) \\ \Rightarrow m^2-n^2=\frac{co{t}^2\theta .\sin \theta }{4} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides}, \\ {\left(m^2-n^2\right)}^2=\frac{co{t}^4\theta .{\sin }^2\theta }{16} \\ \Rightarrow {\left(m^2-n^2\right)}^2=\frac{{\cos }^4\theta }{16{\sin }^2\theta }\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right): \\ {\left({\mathrm{m}}^2-{\mathrm{n}}^2\right)}^2=\mathrm{m}\mathrm{n} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$