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Inverse Trigonometric Functions question

2023 · 13 Apr · Shift 1 · Q41
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Inverse Trigonometric Functions question

2023 · 13 Apr · Shift 1 · Q41

JEE MainMathematicsInverse Trigonometric FunctionsNumerical+4 / −1
If S={x∈R:sin⁡−1(x+1x2+2x+2)−sin⁡−1(xx2+1)=π4}S=\left\{x \in \mathbb{R}: \sin ^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}\right\}S={x∈R:sin−1(x2+2x+2​x+1​)−sin−1(x2+1​x​)=4π​}, then ∑x∈s(sin⁡((x2+x+5)π2)−cos⁡((x2+x+5)π))\sum_{x \in s}\left(\sin \left(\left(x^{2}+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^{2}+x+5\right) \pi\right)\right)x∈s∑​(sin((x2+x+5)2π​)−cos((x2+x+5)π)) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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