Question

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

Hint:

Use trigonometric identities to simplify inverse trigonometric functions. In this case, $\sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right)$ simplifies to $\tan^{-1}(x)$.

Answer 1

We are asked to simplify $\sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right)$. Let: \[ \theta = \sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right). \] Thus, \[ \sin(\theta) = \frac{x}{\sqrt{1 + x^2}}. \] Now, to find $\theta$, we can use the identity: \[ \cos^2(\theta) + \sin^2(\theta) = 1. \] Substitute $\sin(\theta) = \frac{x}{\sqrt{1 + x^2}}$: \[ \cos^2(\theta) + \left( \frac{x}{\sqrt{1 + x^2}} \right)^2 = 1 \quad \Rightarrow \quad \cos^2(\theta) = 1 - \frac{x^2}{1 + x^2}. \] Simplifying the right-hand side: \[ \cos^2(\theta) = \frac{1 + x^2 - x^2}{1 + x^2} = \frac{1}{1 + x^2}. \] Thus, \[ \cos(\theta) = \frac{1}{\sqrt{1 + x^2}}. \] Therefore, \[ \theta = \tan^{-1}(x). \] Thus, \[ \sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right) = \tan^{-1}(x). \]