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### JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
If $$\alpha$$ $$= 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right)$$ and $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right),$$ where the inverse trigonimetric functions take only the principal values, then the correct options(s) is (are)
A
$$cos\beta > 0$$
B
$$\sin \beta < 0$$
C
$$\cos \left( {\alpha + \beta } \right) > 0$$
D
$$\cos \alpha < 0$$

## 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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