Question

If $x=3{\cos }^{-1}\left(\frac{5}{11}\right)$ and $y=3{\sin }^{-1}\left(\frac{7}{13}\right),$ where the inverse trigonometric functions take only the principal values, then the CORRECT option is/are

Answer 1

$\because {\cos }^{-1}x$ is decreasing function
$\therefore x=3{\cos }^{-1}\left(\frac{5}{11}\right)>3{\cos }^{-1}\left(\frac{5}{10}\right)=3{\cos }^{-1}\left(\frac{1}{2}\right)=3\times \frac{\pi }{3}=\pi$
$\Rightarrow x>\pi$

and $x=3{\cos }^{-1}\left(\frac{5}{11}\right)<3{\cos }^{-1}\left(0\right)=\frac{3\pi }{2}$
$\Rightarrow x<\frac{3\pi }{2}$
$\Rightarrow \pi <x<\frac{3\pi }{2}$
$\sin x<0$ and $\cos x<0\dots \left(1\right)$

And
$\because {\sin }^{-1}x$ is increasing function
$\therefore y=3{\sin }^{-1}\left(\frac{7}{13}\right)>3{\sin }^{-1}\left(\frac{1}{2}\right)=3\times \frac{\pi }{6}=\frac{\pi }{2}$
$\Rightarrow y>\frac{\pi }{2}$

And $y=3{\sin }^{-1}\left(\frac{7}{13}\right)<3{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)=3\times \frac{\pi }{3}=\pi$
$\Rightarrow y<\pi$
$\Rightarrow \frac{\pi }{2}<y<\pi \dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
And $\frac{3\pi }{2}<x+y<\frac{5\pi }{2}$
$\Rightarrow \cos (x+y)>0$
and $\sin (x+y)>0$ is not always true