Question

Write \(\tan^{-1}\left(\dfrac{x}{\sqrt{a^2-x^2}}\right),\ |x|<a\) in simplest form.

Hint:

Substitute \(x=a\sin\theta\) to clear the square root, then simplify \(\tan^{-1}(\tan\theta)\).

Answer 1

Step 1: Understanding the Concept:
The presence of \(\sqrt{a^2-x^2}\) suggests the substitution \(x=a\sin\theta\), which is the standard trick that clears this kind of square root.

Step 2: Substituting:
Let \(x=a\sin\theta\), so \(\theta=\sin^{-1}(x/a)\), and \(\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin^2\theta}=a\cos\theta\) (since \(|x|<a\) keeps \(\cos\theta>0\)).

Step 3: Simplifying the expression:
\(\dfrac{x}{\sqrt{a^2-x^2}}=\dfrac{a\sin\theta}{a\cos\theta}=\tan\theta\), so \(\tan^{-1}\left(\dfrac{x}{\sqrt{a^2-x^2}}\right)=\tan^{-1}(\tan\theta)=\theta\).

Final Answer:
Simplest form \(=\boxed{\sin^{-1}\left(\dfrac{x}{a}\right)}\).