JEE Main 2023 — Inverse Trigonometric Functions Question with Solution

We're given the equation ${\sin }^{-1}x=2{\tan }^{-1}x$ for $x$ in the interval $(-1,1]$. We want to find the number of solutions.

Step 1: Apply the sine and tangent functions to both sides :

We can rewrite the equation by applying the sine function to both sides :

$$\sin ({\sin }^{-1}x)=\sin (2{\tan }^{-1}x).$$

This simplifies to:

$$x=\sin (2{\tan }^{-1}x).$$

Step 2: Use the double-angle identity for sine :

Recall that $\sin (2y)=2\sin (y)\cos (y)$. Applying this identity to the right-hand side gives :

$$x=2\sin ({\tan }^{-1}x)\cos ({\tan }^{-1}x).$$

Step 3: Use the identities for sine and cosine of an inverse tangent :

Recall that $\sin ({\tan }^{-1}x)=\frac{x}{\sqrt{1+x^2}}$ and $\cos ({\tan }^{-1}x)=\frac{1}{\sqrt{1+x^2}}$. Substituting these into the equation gives :

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

This simplifies to :

$$x=\frac{2x}{1+x^2}.$$

Step 4: Solve for $x$ :

We have :

$$x=\frac{2x}{1+x^2}.$$

Cross-multiplying gives :

$$x(1+x^2)=2x.$$

This simplifies to :

$$x^3+x-2x=0.$$

Rearranging terms gives :

$$x^3-x=0.$$

This factors to:

$$x(x^2-1)=0.$$

Setting each factor equal to zero gives the solutions $x=0$, $x=-1$, and $x=1$.

However, we are given that $x\in (-1,1]$. Therefore, the only solutions in this interval are $x=0$ and $x=1$.

So there are 2 solutions to the equation ${\sin }^{-1}x=2{\tan }^{-1}x$ in the interval $x\in (-1,1]$.