Question

If α + β − γ = π and sin² α +sin² β − sin² γ = λ sin α sin β cos γ, then write the value of λ.

Answer 1

$$\begin{gathered} \text{Given:} \\ \gamma =-\left[\mathrm{\pi }-(\mathrm{\alpha }+\mathrm{\beta })\right] \\ \mathrm{Also}, \\ \lambda =\frac{{\sin }^2\alpha +{\sin }^2\beta -{\sin }^2\left[-(\mathrm{\pi }-(\mathrm{\alpha }+\mathrm{\beta })\right]}{\sin \alpha \sin \beta \cos (-(\mathrm{\pi }-(\mathrm{\alpha }+\mathrm{\beta }))} \\ =\frac{{\sin }^2\alpha +{\sin }^2\beta -\left(\sin \right(\alpha +\beta ){)}^2}{-(\sin \alpha \sin \beta \cos (\alpha +\beta \left)\right)}\left[\sin \left(\mathrm{\pi }-\mathrm{\theta }\right)=\sin \theta \mathrm{and}\cos \left(\mathrm{\pi }-\mathrm{\theta }\right)=-\cos \theta \right] \\ =\frac{{\sin }^2\alpha +{\sin }^2\beta -{\sin }^2\alpha {\cos }^2\beta -{\cos }^2\alpha {\sin }^2\beta -2\sin \alpha \sin \beta \cos \alpha \cos \beta }{-(\sin \alpha \sin \beta \cos \alpha \cos \beta -{\sin }^2\alpha {\sin }^2\beta )} \\ =\frac{{\sin }^2\alpha (1-{\cos }^2\beta )+{\sin }^2\beta (1-{\cos }^2\alpha )-2\sin \alpha \sin \beta \cos \alpha \cos \beta }{{\sin }^2\alpha {\sin }^2\beta -\sin \alpha \sin \beta \cos \alpha \cos \beta } \\ =\frac{2{\sin }^2\alpha {\sin }^2\beta -2\sin \alpha \sin \beta \cos \alpha \cos \beta }{{\sin }^2\alpha {\sin }^2\beta -\sin \alpha \sin \beta \cos \alpha \cos \beta } \\ =2 \end{gathered}$$