Q.$\sin (\tan^{-1} x)$, $|x|<1$ is equal to (A) $\frac{x}{\sqrt{1-x^2}}$ (B) $\frac{1}{\sqrt{1-x^2}}$ (C) $\frac{1}{\sqrt{1+x^2}}$ (D) $\frac{x}{\sqrt{1+x^2}}$

To simplify $\sin(\tan^{-1} x)$, name the inverse as an angle and rebuild the missing ratio from a right triangle.

Let $\theta = \tan^{-1} x$, so $\tan \theta = x$ with $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Since $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{1}$, take opposite $= x$ and adjacent $= 1$; by Pythagoras the hypotenuse is $\sqrt{1+x^2}$. …

The key here is to avoid memorising formulas blindly. Instead, think of $\tan^{-1} x$ as an angle — call it $\theta$ — whose tangent is $x$. Then the problem becomes: find $\sin \theta$ given that $\tan \theta = x$. That’s a pure trigonometry problem, and the neatest way is to draw a right triangle.

1. Set up the angle.

   Let $\theta = \tan^{-1} x$. By definition, $\tan \theta = x$ and $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Since $|x| < 1$, $\theta$ is a small angle, but that doesn’t affect the algebra.

2. Build the triangle.

   For $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{1}$, we can take the opposite side as $x$ and the adjacent side as $1$.

   The hypotenuse, by Pythagoras, is $\sqrt{x^2 + 1^2} = \sqrt{x^2 + 1}$.

   Watch out

   A common mistake is to write the hypotenuse as $\sqrt{1 - x^2}$ — that would be for $\sin^{-1} x$ or $\cos^{-1} x$, where the ratio is between sides and the hypotenuse. Here the ratio is between the two legs, so it’s $1 + x^2$ under the root, not $1 - x^2$.

3. Read off $\sin \theta$.

   $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{1 + x^2}}$. …

Method: Simplifying a trig function of an inverse trig function

Use this for compositions like $\sin(\tan^{-1}x)$, $\cos(\tan^{-1}x)$, or $\tan(\sin^{-1}x)$.

Steps

Step 1: Let the inverse part be an angle.

Set $\theta = \tan^{-1}x$, so $\tan\theta = x$ and $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Step 2: Build the right triangle from the known ratio. …

Common Mistakes

Mistake 1: Using $\sqrt{1-x^2}$ for the hypotenuse.

Why it's wrong: with $\tan\theta = \frac{x}{1}$ the ratio is between the two legs, so the hypotenuse is $\sqrt{1+x^2}$; $\sqrt{1-x^2}$ belongs to $\sin^{-1}$/$\cos^{-1}$ setups. Correct approach: hypotenuse $= \sqrt{x^2 + 1}$, giving $\sin\theta = \frac{x}{\sqrt{1+x^2}}$ (option D).

Mistake 2: Picking option (A) $\frac{x}{\sqrt{1-x^2}}$ by confusing it with $\tan(\sin^{-1}x)$. …