Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given, ${\sin }^{-1}\left(\sin \theta \right)-{\cos }^{-1}\left(\sin \theta \right)>0,\theta \in \left(0,2\pi \right)$

$\because {\sin }^{-1}\left(\theta \right)+{\cos }^{-1}\left(\theta \right)=\frac{\mathrm{\pi }}{2}$

$\therefore {\sin }^{-1}\left(\sin \theta \right)-\left(\frac{\pi }{2}-{\sin }^{-1}\left(\sin \theta \right)\right)>0$

${\sin }^{-1}\left(\sin \theta \right)>\frac{\pi }{4}\Rightarrow \theta \in \left(\frac{\pi }{4},\frac{3\pi }{4}\right)$

$\left(a,b\right)=\left(\frac{\pi }{4},\frac{3\pi }{4}\right)\Rightarrow b-a=\frac{\pi }{2}$

Given $b-a=\alpha -\beta$

$\therefore \alpha -\beta =\frac{\mathrm{\pi }}{2} .....\left(1\right)$

Now $\alpha x^2+\beta x+{\sin }^{-1}\left(x^2-6x+10\right)+{\cos }^{-1}\left(x^2-6x+10\right)=0$

Now defining $x^2-6x+10=1+{\left(x-3\right)}^2\ge 1$

Hence $x=3$ is the only possible solution

$9\alpha +3\beta +\frac{\pi }{2}=0 \dots \left(2\right)$

On solving equations $\left(1\right)$ and $\left(2\right)$ we get,

$\alpha =\frac{\pi }{12}$