Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

${\left({\sin }^{-1}x\right)}^2-{\left({\cos }^{-1}x\right)}^2=a$

Let ${\sin }^{-1}\left(x\right)=t$

$x=\sin t$

${\left({\sin }^{-1}\sin t\right)}^2-{\left({\cos }^{-1}\sin t\right)}^2=a$

$t^2-{\left(\frac{\pi }{2}-t\right)}^2=a$

$t^2-\left(\frac{{\pi }^2}{4}+t^2-\pi t\right)=a$

$\pi t-\frac{{\pi }^2}{4}=a$

$t=\frac{a}{\pi }+\frac{\pi }{4}$

$x=\sin \left(\frac{a}{\pi }+\frac{\pi }{4}\right)$

$2x^2-1=2{\sin }^2\left(\frac{a}{\pi }+\frac{\pi }{4}\right)-1$

$=2{\left(\sin \frac{a}{\pi } \cos \frac{\pi }{4}+\cos \frac{a}{\pi } \sin \frac{\pi }{4}\right)}^2-1$

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

$={\sin }^2\frac{a}{\pi }+{\cos }^2\frac{\mathit{a}}{\mathit{\pi }}+2\cos \frac{a}{\pi }\sin \frac{a}{\pi }-1$

$=2\cos \frac{a}{\pi }\sin \frac{a}{\pi }$

$=\sin \frac{2a}{\pi }$