JEE Main 2023 — Mathematics Trigonometry

Given,

$${\mathrm{sin}}^{-1}(\frac{x+1}{\sqrt{{x}^{2}+2x+2}})-{\mathrm{sin}}^{-1}(\frac{x}{\sqrt{{x}^{2}+1}})=\frac{\pi }{4}$$sin−1(x2+2x+2​x+1​)−sin−1(x2+1​x​)=4π​

$$\Rightarrow {\mathrm{sin}}^{-1}(\frac{x+1}{\sqrt{{x}^{2}+x+2}})=\frac{\pi }{4}+{\mathrm{sin}}^{-1}\frac{x}{\sqrt{{x}^{2}+1}}$$⇒sin−1(x2+x+2​x+1​)=4π​+sin−1x2+1​x​

$$\Rightarrow (\frac{x+1}{\sqrt{{x}^{2}+x+2}})=\mathrm{sin}(\frac{\pi }{4}+{\mathrm{sin}}^{-1}\frac{x}{\sqrt{{x}^{2}+1}})$$⇒(x2+x+2​x+1​)=sin(4π​+sin−1x2+1​x​)

$$\Rightarrow (\frac{x+1}{\sqrt{{x}^{2}+x+2}})=\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{{x}^{2}+1}}+\frac{1}{\sqrt{2}}\times \frac{x}{\sqrt{{x}^{2}+1}}$$⇒(x2+x+2​x+1​)=2​1​×x2+1​1​+2​1​×x2+1​x​

$$\Rightarrow \frac{x+1}{\sqrt{{x}^{2}+x+2}}=\frac{x+1}{\sqrt{2}\sqrt{{x}^{2}+1}}$$⇒x2+x+2​x+1​=2​x2+1​x+1​

$$\Rightarrow (x+1)(\sqrt{2}\sqrt{{x}^{2}+1}-\sqrt{{x}^{2}+x+2})=0$$⇒(x+1)(2​x2+1​−x2+x+2​)=0

$\Rightarrow x=-1$⇒x=−1 or $\sqrt{{x}^{2}+x+2}=\sqrt{2}\cdot \sqrt{{x}^{2}+1}$x2+x+2​=2​⋅x2+1​

Now solving, $\sqrt{{x}^{2}+x+2}=\sqrt{2}\cdot \sqrt{{x}^{2}+1}$x2+x+2​=2​⋅x2+1​ we get,

$${x}^{2}+x+2=2({x}^{2}+1)$$x2+x+2=2(x2+1)

$$\Rightarrow {x}^{2}-x=0$$⇒x2−x=0

$\Rightarrow x=0,x=1$⇒x=0,x=1 {rejected as $x=1$x=1 will not satisfy the given equation}

Hence, $S={0,1}$S=0,1

Now solving,

$$\underset{n\in S}{\sum }(\mathrm{sin}({x}^{2}+x+5)\frac{\pi }{2}-\mathrm{cos}({x}^{2}+x+5)\pi )$$n∈S∑​(sin(x2+x+5)2π​−cos(x2+x+5)π)

$$=(\mathrm{sin}\frac{5\pi }{2}-\mathrm{cos}5\pi +\mathrm{sin}\frac{5\pi }{2}-\mathrm{cos}5\pi )$$=(sin25π​−cos5π+sin25π​−cos5π)

$$=1-(-1)+1-(-1)=4$$=1−(−1)+1−(−1)=4