JEE Main 2023 — Inverse Trigonometric Functions Question with Solution

Given,

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

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

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

$\Rightarrow \left(\frac{x+1}{\sqrt{x^2+x+2}}\right)=\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{x^2+1}}+\frac{1}{\sqrt{2}}\times \frac{x}{\sqrt{x^2+1}}$

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

$\Rightarrow \left(x+1\right)\left(\sqrt{2}\sqrt{x^2+1}-\sqrt{x^2+x+2}\right)=0$

$\Rightarrow x=-1$ or $\sqrt{x^2+x+2}=\sqrt{2}·\sqrt{x^2+1}$

Now solving, $\sqrt{x^2+x+2}=\sqrt{2}·\sqrt{x^2+1}$ we get,

$x^2+x+2=2\left(x^2+1\right)$

$\Rightarrow x^2-x=0$

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

Hence, $S=\left\{0,1\right\}$

Now solving,

$\sum _{n\in S}\left(\sin \left(x^2+x+5\right)\frac{\pi }{2}-\cos \left(x^2+x+5\right)\pi \right)$

$=\left(\sin \frac{5\pi }{2}-\cos 5\pi +\sin \frac{5\pi }{2}-\cos 5\pi \right)$

$=1-\left(-1\right)+1-\left(-1\right)=4$