The-solution-of-sin-1-2a-1-a-2-cos-1-1-b-2-1-b-2-tan-1-2x-1-x-2-is-

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=tanα b=tanβ x=tanγ sin^(−1) (((2tanα)/(1+tan^2 α)))−cos^(−1) (((1−tan^2 β)/(1+tan^2 β)))=tan^(−1) (((2tanγ)/(1−tan^2 γ))) sin^(−1) (sin2α)−cos^(−1) (cos2β)=tan^(−1) (tan2γ) 2α−2β=2γ α−β=γ tan(α−β)=tanγ ((tanα−tanβ)/(1+tanαtanβ))=tanγ ((a−b)/(1+ab))=x

$${a}={tan}\alpha\:\:\:\:{b}={tan}\beta\:\:\:\:{x}={tan}\gamma \\ $$$${sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{tan}\alpha}{\mathrm{1}+{tan}^{\mathrm{2}} \alpha}\right)−{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{tan}^{\mathrm{2}} \beta}{\mathrm{1}+{tan}^{\mathrm{2}} \beta}\right)={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{tan}\gamma}{\mathrm{1}−{tan}^{\mathrm{2}} \gamma}\right) \\ $$$${sin}^{−\mathrm{1}} \left({sin}\mathrm{2}\alpha\right)−{cos}^{−\mathrm{1}} \left({cos}\mathrm{2}\beta\right)={tan}^{−\mathrm{1}} \left({tan}\mathrm{2}\gamma\right) \\ $$$$\mathrm{2}\alpha−\mathrm{2}\beta=\mathrm{2}\gamma \\ $$$$\alpha−\beta=\gamma \\ $$$${tan}\left(\alpha−\beta\right)={tan}\gamma \\ $$$$\frac{{tan}\alpha−{tan}\beta}{\mathrm{1}+{tan}\alpha{tan}\beta}={tan}\gamma \\ $$$$\frac{{a}−{b}}{\mathrm{1}+{ab}}={x} \\ $$$$ \\ $$

Commented by gunawan last updated on 08/Sep/18