To explain why $\sin^{-1}(t) = \tan^{-1}\left(\frac{t}{\sqrt{1-t^2}}\right)$ when $-1 < t < 1$, we can derive this relationship mathematically.

Step-by-Step Explanation:

1. Set the definition of $\sin^{-1}(t)$: Let $\theta = \sin^{-1}(t)$, where $-1 \leq t \leq 1$. By the definition of inverse sine: $\sin(\theta) = t \quad \text{and} \quad -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}.$

2. Use the Pythagorean identity: From $\sin^2(\theta) + \cos^2(\theta) = 1$, we know: $\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - t^2},$ where the positive root is taken because $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, and cosine is non-negative in this interval.

3. Express $\tan(\theta)$: Recall that: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.$ Substituting $\sin(\theta) = t$ and $\cos(\theta) = \sqrt{1 - t^2}$, we get: $\tan(\theta) = \frac{t}{\sqrt{1 - t^2}}.$

4. Relate $\theta$ to $\tan^{-1}$: By the definition of the inverse tangent function, $\theta = \tan^{-1}\left(\frac{t}{\sqrt{1 - t^2}}\right)$.

5. Combine results: Since $\theta = \sin^{-1}(t)$ and $\theta = \tan^{-1}\left(\frac{t}{\sqrt{1 - t^2}}\right)$, it follows that: $\sin^{-1}(t) = \tan^{-1}\left(\frac{t}{\sqrt{1 - t^2}}\right),$ provided that $-1 < t < 1$.

Why the Range Matters:

The restriction $-1 < t < 1$ ensures that:

• $\sqrt{1 - t^2}$ is real and positive.
• The domains of both $\sin^{-1}(t)$ and $\tan^{-1}(x)$ are valid for real values.

Would you like further details or graphical illustrations of this explanation?

Follow-Up Questions:

1. Why does the Pythagorean identity play a key role in trigonometric inverses?
2. How can this relationship be visualized on a unit circle?
3. What happens to the equation when $t = \pm1$?
4. How would this equation be derived if $\cos^{-1}(t)$ were involved instead of $\sin^{-1}(t)$?
5. Can this be extended to complex values of $t$?

Tip:

Always cross-check inverse trigonometric relationships by testing with specific values within the allowed range!