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: Alternative substitution:
Let \(x=a\cos\phi\) instead; then \(\sqrt{a^2-x^2}=a\sin\phi\), and \(\dfrac{x}{\sqrt{a^2-x^2}}=\dfrac{a\cos\phi}{a\sin\phi}=\cot\phi=\tan\left(\dfrac{\pi}{2}-\phi\right)\).

Step 2: Converting back:
So the expression equals \(\dfrac{\pi}{2}-\phi=\dfrac{\pi}{2}-\cos^{-1}(x/a)\), which by the complementary identity \(\dfrac{\pi}{2}-\cos^{-1}t=\sin^{-1}t\) is exactly \(\sin^{-1}(x/a)\).

Final Answer:
Same result: \(\boxed{\sin^{-1}(x/a)}\).