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Question Number 197937 by mathlove last updated on 05/Oct/23

$$\sqrt{3}{sin}^2\theta \cdot tan\beta +{cos}^2\beta =?$$

Answered by Frix last updated on 05/Oct/23

$$\mathrm{It}\mathrm{is}\mathrm{what}\mathrm{it}\mathrm{is},\mathrm{not}\mathrm{getting}\mathrm{better}\mathrm{any}\mathrm{way}.$$$$p=\tan \theta \wedge q=\tan \beta \to \frac{\sqrt{3}{p}^{2}q}{p^2+1}+\frac{1}{q^2+1}$$

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