Question

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

Answer 1

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