Inverse Trigonometric Functions

$a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots .$. be consecutive natural numbers.

$ \begin{aligned} & \therefore \quad a_2=2, a_3=3, \ldots ., a_{2021}=2021, a_{2022}=2022 \\\\ & \tan ^{-1}\left[\frac{1}{1+a_1 a_2}\right]=\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2}\right) \\\\ & \tan ^{-1}\left[\frac{1}{1+a_2 a_3}\right]=\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right]=\tan ^{-1}\left(\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{3}\right) \end{aligned} $

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$ \begin{aligned} \tan ^{-1}\left[\frac{1}{1+a_{2021} a_{2022}}\right] & =\tan ^{-1}\left[\frac{1}{1+2021 \cdot 2022}\right] \\\\ & =\tan ^{-1}\left(\frac{1}{2021}\right)-\tan ^{-1}\left(\frac{1}{2022}\right) \end{aligned} $

$ \therefore $ $\tan ^{-1}\left(\frac{1}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+a_{2} a_{3}}\right)+\ldots . .+\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)$

$ =\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2022}\right)=\frac{\pi}{4}-\cot ^{-1}(2022) $

$ =\frac{\pi}{4}-\left(\frac{\pi}{2}-\tan ^{-1}(2022)\right)=\tan ^{-1}(2022)-\frac{\pi}{4} $

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 ${{3\pi } \over 2}$.

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