JEE Advance - Mathematics (2015 - Paper 2 Offline - No. 12)
Explanation
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 $${{3\pi } \over 2}$$.
$$\therefore$$ $$\cos (\alpha + \beta ) > 0$$
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