Question Number 199817 by ajfour last updated on 09/Nov/23
Answered by ajfour last updated on 09/Nov/23
$$\left\{{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta\right)−\frac{\mathrm{sin}\:\theta}{\mathrm{2}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)}\right\}^{\mathrm{2}} \\ $$$$\:\:\:={a}\left\{{a}+\frac{\mathrm{sin}\:\theta}{\mathrm{2cos}\:\theta\left(\mathrm{1}+\mathrm{sin}\:\theta\right)}\right\} \\ $$$${b}=\frac{\mathrm{sin}\:\theta}{\mathrm{2cos}\:\theta\left(\mathrm{1}+\mathrm{sin}\:\theta\right)} \\ $$
Answered by ajfour last updated on 09/Nov/23
