Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

$\alpha =3{\sin }^{-1}\frac{6}{11}$
Since$\frac{\pi }{4}>{\sin }^{-1}\frac{6}{11}>\frac{\pi }{6}$
$\frac{3\pi }{4}>3{\sin }^{-1}\frac{6}{11}>\frac{\pi }{2}$
$\cos \alpha <0$
$\cos \left(3{\cos }^{-1}\frac{4}{9}\right)=4\times {\left(\frac{4}{9}\right)}^3-3\times \frac{4}{9}$
$=\frac{4\times 64}{729}-\frac{12}{9}$
$=\frac{256-12\times 81}{72}<0$
$\frac{5\pi }{12}>{\cos }^{-1}\frac{4}{9}>\frac{\pi }{3}$
$\frac{5\pi }{4}>3{\cos }^{-1}\frac{4}{9}>\pi$
$\sin \left(3\frac{co{s}^{-1}4}{9}\right)<0$
$\frac{3\pi }{4}>3{\sin }^{-1}\frac{6}{11}>\frac{\pi }{2}$
$\frac{5\pi }{4}>3{\cos }^{-1}\frac{4}{9}>\pi$
$2\pi >3{\sin }^{-1}\frac{6}{11}+3{\cos }^{-1}\frac{4}{9}>\frac{3\pi }{2}$
$\cos \left(3{\sin }^{-1}\frac{6}{11}+3{\cos }^{-1}\frac{4}{9}\right)>0$