If S = {x in R : sin^{-1}left(frac{x+1}{sqrt{ | vedclass

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(C) Let $f(x) = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}$.Let $\alpha = \sin^

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$4$

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(C) Let $f(x) = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}$.Let $\alpha = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right)$ and $\beta = \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)$.Then $\sin \alpha = \frac{x+1}{\sqrt{(x+1)^2+1}}$ and $\sin \beta = \frac{x}{\sqrt{x^2+1}}$.Since $\cos \alpha = \frac{1}{\sqrt{(x+1)^2+1}}$ and $\cos \beta = \frac{1}{\sqrt{x^2+1}}$,we have $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = \frac{x+1}{\sqrt{(x+1)^2+1}}\frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt{(x+1)^2+1}}\frac{x}{\sqrt{x^2+1}} = \frac{1}{\sqrt{(x+1)^2+1}\sqrt{x^2+1}}$.Given $\alpha - \beta = \frac{\pi}{4}$,so $\sin(\alpha - \beta) = \frac{1}{\sqrt{2}}$.Thus,$\frac{1}{\sqrt{(x+1)^2+1}\sqrt{x^2+1}} = \frac{1}{\sqrt{2}} \implies ((x+1)^2+1)(x^2+1) = 2$.$(x^2+2x+2)(x^2+1) = 2 \implies x^4 + 2x^3 + 3x^2 + 2x + 2 = 2 \implies x(x^3 + 2x^2 + 3x + 2) = 0$.$x(x+1)(x^2+x+2) = 0$. Since $x^2+x+2$ has no real roots,$x=0$ or $x=-1$.For $x=0$,$\sin^{-1}(1/\sqrt{2}) - \sin^{-1}(0) = \pi/4 - 0 = \pi/4$ (Valid).For $x=-1$,$\sin^{-1}(0) - \sin^{-1}(-1/\sqrt{2}) = 0 - (-\pi/4) = \pi/4$ (Valid).So $S = \{-1, 0\}$.For $x=0$,$\sin(5\pi/2) - \cos(5\pi) = 1 - (-1) = 2$.For $x=-1$,$\sin(5\pi/2) - \cos(5\pi) = 1 - (-1) = 2$.Sum $= 2 + 2 = 4$.

If $S = \{x \in R : \sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}\}$,then $\sum_{x \in S} \left(\sin\left((x^2+x+5)\frac{\pi}{2}\right) - \cos((x^2+x+5)\pi)\right)$ is equal to $........$.

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