Question Number 202295 by liuxinnan last updated on 24/Dec/23
$${psin}\theta{con}^{\mathrm{2}} \theta={a} \\ $$$${pcos}\theta{sin}^{\mathrm{2}} \theta={b} \\ $$$${p}\neq\mathrm{0}\:\:\:\:\theta\in\left(\mathrm{0}\:,\frac{\pi}{\mathrm{2}}\right) \\ $$$${prove}\:{p}=\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{ab}} \\ $$
Answered by 1990mbodji last updated on 24/Dec/23