Question Number 43179 by gunawan last updated on 08/Sep/18
$$\mathrm{The}\:\mathrm{solution}\:\mathrm{of} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{a}}{\mathrm{1}+{a}^{\mathrm{2}} }\right)−\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{b}^{\mathrm{2}} }{\mathrm{1}+{b}^{\mathrm{2}} }\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)\:\mathrm{is} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Sep/18
$${a}={tan}\alpha\:\:\:\:{b}={tan}\beta\:\:\:\:{x}={tan}\gamma \\ $$$${sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{tan}\alpha}{\mathrm{1}+{tan}^{\mathrm{2}} \alpha}\right)−{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{tan}^{\mathrm{2}} \beta}{\mathrm{1}+{tan}^{\mathrm{2}} \beta}\right)={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{tan}\gamma}{\mathrm{1}−{tan}^{\mathrm{2}} \gamma}\right) \\ $$$${sin}^{−\mathrm{1}} \left({sin}\mathrm{2}\alpha\right)−{cos}^{−\mathrm{1}} \left({cos}\mathrm{2}\beta\right)={tan}^{−\mathrm{1}} \left({tan}\mathrm{2}\gamma\right) \\ $$$$\mathrm{2}\alpha−\mathrm{2}\beta=\mathrm{2}\gamma \\ $$$$\alpha−\beta=\gamma \\ $$$${tan}\left(\alpha−\beta\right)={tan}\gamma \\ $$$$\frac{{tan}\alpha−{tan}\beta}{\mathrm{1}+{tan}\alpha{tan}\beta}={tan}\gamma \\ $$$$\frac{{a}−{b}}{\mathrm{1}+{ab}}={x} \\ $$$$ \\ $$
Commented by gunawan last updated on 08/Sep/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 08/Sep/18
$${you}\:{are}\:{welcome}...{everybody}\:{is}\:{uploading}\:{self} \\ $$$${photo}...{so}\:{you}\:{pls}... \\ $$
Commented by gunawan last updated on 08/Sep/18
Commented by tanmay.chaudhury50@gmail.com last updated on 08/Sep/18
$${excellent}... \\ $$