Simplifying $\sin(\tan^{-1}x)$

The question asks us to find the value of the expression $\sin(\tan^{-1}x)$ under the condition that $|x| < 1$. This involves understanding inverse trigonometric functions and their relationship with standard trigonometric functions.

Understanding Inverse Tangent

Let $y = \tan^{-1}x$. By definition, this means that $\tan y = x$. The condition $|x| < 1$ implies that $y$ lies in the interval $(-\frac{\pi}{4}, \frac{\pi}{4})$.

Using Trigonometric Identities

We need to find $\sin y$ given that $\tan y = x$. We can use the fundamental trigonometric identity relating tangent and sine:

$ \tan y = \frac{\sin y}{\cos y} $

So, we have:

$ x = \frac{\sin y}{\cos y} $

We also know the Pythagorean identity:

$ \sin^2 y + \cos^2 y = 1 $

From $x = \frac{\sin y}{\cos y}$, we get $\cos y = \frac{\sin y}{x}$. Substituting this into the Pythagorean identity:

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

$ \sin^2 y + \frac{\sin^2 y}{x^2} = 1 $

Factor out $\sin^2 y$:

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

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

$ \sin^2 y = \frac{x^2}{x^2 + 1} $

Taking the square root of both sides:

$ \sin y = \pm \sqrt{\frac{x^2}{x^2 + 1}} = \pm \frac{|x|}{\sqrt{x^2 + 1}} $

Since $y = \tan^{-1}x$ and $|x| < 1$, $y$ is in the interval $(-\frac{\pi}{4}, \frac{\pi}{4})$. In this interval, $\sin y$ has the same sign as $y$, which in turn has the same sign as $x$. Therefore, we can replace $\pm |x|$ with $x$.

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

Substituting back $y = \tan^{-1}x$, we get:

$ \sin(\tan^{-1}x) = \frac{x}{\sqrt{1+x^2}} $

Using a Right Triangle Visualization

Another way to solve this is by visualizing a right-angled triangle.

1. Let $y = \tan^{-1}x$. This means $\tan y = x$.
2. We can write $\tan y$ as the ratio $\frac{\text{opposite}}{\text{adjacent}}$. So, we can set the opposite side to $x$ and the adjacent side to $1$ (assuming $x>0$ for simplicity, the result holds for $x<0$ as well).
3. Calculate the hypotenuse using the Pythagorean theorem ($a^2 + b^2 = c^2$):

   Hypotenuse$^2$ = Opposite$^2$ + Adjacent$^2$

   Hypotenuse$^2 = x^2 + 1^2 = x^2 + 1$

   Hypotenuse = $\sqrt{x^2 + 1}$

4. Now, find $\sin y$, which is the ratio $\frac{\text{opposite}}{\text{hypotenuse}}$:

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

This method confirms the result obtained using identities.

Conclusion

Therefore, for $|x| < 1$, the expression $\sin(\tan^{-1}x)$ simplifies to $\frac{x}{\sqrt{1+x^2}}$.