If $x = c o s^{2} (t a n^{- 1} (s i n (c o t^{- 1} 3)))$ , then $1331 x^{3} - 3630 x^{2} + 3300 x + 7369 = m$ then find the sum of the second and third digits of $m$

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Solution

$x = c o s^{2} (t a n^{- 1} (s i n (c o t^{- 1} 3)))$
$\Rightarrow x = c o s^{2} (t a n^{- 1} (s i n (s i n^{- 1} (1 / \sqrt{10}))))$
$\Rightarrow x = c o s^{2} (t a n^{- 1} (1 / \sqrt{10}))$
$\Rightarrow x = c o s^{2} (c o s^{- 1} (\sqrt{10 / 11}))$
$\Rightarrow x = 10 / 11$
So,
$1331 x^{3} - 3630 x^{2} + 3300 x + 7369 = m$ (Given)
$= (11 x - 10)^{3} + 8369 = 8369$ as $11 x - 10 = 0$
Hence $m = 8369$
Sum of second and third digits of $m$ is $= 3 + 6 = 9$

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