Inverse Trigonometric Functions
Substitution
Premium Question
Grade 12
Question:
For $|x| < 1$, $x \neq 0$: $\tan^{-1}\left[\frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}\right]$ equals:
(1) $\frac{\pi}{4} + \frac{1}{2} \cos^{-1} x^2$
(2) $\frac{\pi}{4} + \cos^{-1} x^2$
(3) $\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x^2$
(4) $\frac{\pi}{4} - \cos^{-1} x^2$
Step-by-Step Solution
Key Concept: Let $x^2 = \cos 2A$, so $\sqrt{1 \pm x^2}$ become $\sqrt{2} \cos A$ and $\sqrt{2} \sin A$. The fraction simplifies to $\tan^{-1}(\cot A) = \pi/2 - A$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)