Question Number 197937 by mathlove last updated on 05/Oct/23
$$\sqrt{\mathrm{3}}\:{sin}^{\mathrm{2}} \theta\centerdot{tan}\beta+{cos}^{\mathrm{2}} \beta=? \\ $$
Answered by Frix last updated on 05/Oct/23
$$\mathrm{It}\:\mathrm{is}\:\mathrm{what}\:\mathrm{it}\:\mathrm{is},\:\mathrm{not}\:\mathrm{getting}\:\mathrm{better}\:\mathrm{any}\:\mathrm{way}. \\ $$$${p}=\mathrm{tan}\:\theta\:\wedge{q}=\mathrm{tan}\:\beta\:\rightarrow\:\frac{\sqrt{\mathrm{3}}{p}^{\mathrm{2}} {q}}{{p}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{q}^{\mathrm{2}} +\mathrm{1}} \\ $$