QJEEAdv15Mae240: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Mae240 Solution: If and where the inverse trigonimetric functions take only the…

Here, $\alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right)$

and $\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$ as {6 \over {11}} > {1 \over 2}

\Rightarrow {\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\sin ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 6}

$\therefore$ \alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\pi \over 2} \Rightarrow \cos \alpha < 0

Now, $\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$

As {4 \over 9} < {1 \over 2} \Rightarrow {\cos ^{ - 1}}\left( {{4 \over 9}} \right) > {\cos ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 3}

$\therefore$ \beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right) > \pi

$\therefore$ \cos \beta < 0 and \sin \beta < 0

Now, $\alpha + \beta$ is slightly greater than ${{3\pi } \over 2}$ .

$\therefore$ \cos (\alpha + \beta ) > 0