Question Number 175493 by ajfour last updated on 31/Aug/22
$$\mathrm{tan}^{−\mathrm{1}} \left({a}\mathrm{sin}\:\theta\right)=\mathrm{sin}^{−\mathrm{1}} {b}−\theta \\ $$$${find}\:\theta. \\ $$
Commented by Frix last updated on 01/Sep/22
$$\mathrm{I}\:\mathrm{tried}\:\mathrm{but}\:\mathrm{it}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:\mathrm{4th}\:\mathrm{degree}\:\mathrm{polynome} \\ $$$$\mathrm{which}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{solve} \\ $$
Answered by BHOOPENDRA last updated on 05/Sep/22
$$\theta={b}−\frac{{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} }??? \\ $$