If S = { x : sin - 1 ( x + 1 x 2 + 2 x + 2 ) - sin - 1 ( x x 2 + 1 ) = 4 } , then x S ( sin ( x 2 + x + 5 2 ) - cos ( ( x 2 + x + 5 ) ) ) is equal to _____ . [2023]

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Solution

Q.
If $S = {x \in ℝ : \sin^{- 1} (\frac{x + 1}{\sqrt{x^{2} + 2 x + 2}}) - \sin^{- 1} (\frac{x}{\sqrt{x^{2} + 1}}) = \frac{π}{4}},$ then $\sum_{x \in S} (\sin ((x^{2} + x + 5) \frac{π}{2}) - \cos ((x^{2} + x + 5) π))$ is equal to _____ . [2023]

Ans.
(4)

$Given, \sin^{- 1} (\frac{x + 1}{\sqrt{x^{2} + 2 x + 2}}) - \sin^{- 1} (\frac{x}{\sqrt{x^{2} + 1}}) = \frac{π}{4}$

$\Rightarrow \sin^{- 1} (\frac{x + 1}{\sqrt{(x + 1)^{2} + 1}}) - \sin^{- 1} (\frac{x}{\sqrt{x^{2} + 1}}) = \frac{π}{4}$

$\Rightarrow \tan^{- 1} (x + 1) - \tan^{- 1} (x) = \frac{π}{4} \Rightarrow \tan^{- 1} [\frac{(x + 1) - x}{1 + (x + 1) x}] = \frac{π}{4}$

$\Rightarrow \frac{1}{x^{2} + x + 1} = \tan (\frac{π}{4}) = 1$ $\Rightarrow x^{2} + x + 1 = 1 \Rightarrow x^{2} + x = 0$

$\Rightarrow x = 0 or - 1$ $∴$ $S = {- 1, 0}$

$∴$ $\sum_{x \in S} [\sin ((x^{2} + x + 5) \frac{π}{2}) - \cos ((x^{2} + x + 5) π)]$

$= [\sin (\frac{5 π}{2}) - \cos (5 π)] + {\sin (\frac{5 π}{2}) - \cos (5 π)} = 2 + 2 = 4$

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