JEE Main 2023 — Inverse Trigonometric Functions Question with Solution

Given equation is

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

Let's denote:

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

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

So, we have the equation $A-B=\frac{\pi }{4}$.

We can also write this as $A=B+\frac{\pi }{4}$.

This gives us

$$\sin (A)=\sin \left(B+\frac{\pi }{4}\right).$$

We can use the identity $\sin (a+b)=\sin a\cos b+\cos a\sin b$ and rewrite this equation as:

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

After simplifying, we get:

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

Let's square both sides to remove the square roots:

On the left side, squaring gives:

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

On the right side, squaring gives:

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

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

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

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

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

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

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

$$\begin{array}{rl}& \Rightarrow \frac{x+1}{\sqrt{x^2+2x+2}}=\frac{x+1}{\sqrt{2}\sqrt{x^2+1}}\\ & \\ & \Rightarrow x=-1\text{ OR }\sqrt{x^2+2x+2}=\sqrt{2}\cdot \sqrt{x^2+1}\\ & \\ & \Rightarrow x=0,x=2\text{ (Rejected) }\\ & \\ & S=\{0,-1\}\end{array}$$

$$\begin{array}{rl}& \sum _{x\in R}\left(\sin \left(\left(x^2+x+5\right)\frac{\pi }{2}\right)-\cos \left(\left(x^2+x+5\right)\pi \right)\right)\\ & \\ & =\left[\sin \left(\frac{5\pi }{2}\right)-\cos (5\pi )\right]+\left[\sin \left(\frac{5\pi }{2}\right)-\cos (5\pi )\right]\\ & \\ & =(1-(-1))+(1-(-1))\\ & \\ & =2+2\\ & \\ & =4\end{array}$$