Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given: ${\sin }^{-1}\alpha +{\sin }^{-1}\beta +{\sin }^{-1}\gamma =\pi$

Let, ${\sin }^{-1}\alpha =A, {\sin }^{-1}\beta =B, {\sin }^{-1}\gamma =C$

$\Rightarrow A+B+C=\pi$

Also, $\alpha =\sin A, \beta =\sin B, \gamma =\sin C ...\left(i\right)$

It is given that, ${\left(\alpha +\beta \right)}^2-{\gamma }^2=3\alpha \beta$.

$\Rightarrow {\alpha }^2+{\beta }^2-{\gamma }^2=\alpha \beta$

$\Rightarrow {\sin }^{2}A+{\sin }^{2}B-{\sin }^{2}C=\sin A\sin B$

$\Rightarrow {\sin }^{2}A+\sin \left(B+C\right)\sin \left(B-C\right)=\sin A\sin B$

$\Rightarrow {\sin }^{2}A+\sin \left(\pi -A\right)\sin \left(B-C\right)=\sin A\sin B$

$\Rightarrow {\sin }^{2}A+\sin A\sin \left(B-C\right)=\sin A\sin B$

$\Rightarrow \sin A\left[\sin A+\sin \left(B-C\right)\right]=\sin A\sin B$

$\Rightarrow \sin A\left[\sin A+\sin \left(B-C\right)\right]-\sin A\sin B=0$

$\Rightarrow \sin A\left[\sin A+\sin \left(B-C\right)-\sin B\right]=0$

$\Rightarrow \sin A\left[\sin \left(B+C\right)+\sin \left(B-C\right)-\sin B\right]=0$

$\Rightarrow \sin A\left[2\sin B\cos C-\sin B\right]=0$

$\Rightarrow \sin A\sin B\left[2\cos C-1\right]=0$

$\Rightarrow \sin A=0 \text{or} \sin B=0 \text{or} \cos C=\frac{1}{2}$

It is given that, $\alpha ,\beta ,\gamma \ne 0$

$\Rightarrow \cos C=\frac{1}{2}$

$\Rightarrow \sin C=\frac{\sqrt{3}}{2}$

$\Rightarrow \gamma =\frac{\sqrt{3}}{2}$