Question

If $-1 \leq x \leq 1$, then $\cos^{-1} x + \sin^{-1} x =$

• A. $-\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{2}$
• D. $-\frac{\pi}{16}$

Hint:

Similar identities exist for other pairs: $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ and $\text{cosec}^{-1} x + \sec^{-1} x = \frac{\pi}{2}$. They all stem from the complementary nature of these trigonometric ratios.

Answer 1

The Correct Option is C

This question involves a fundamental identity from inverse trigonometric functions. 1. Theoretical Principle: The identity states that for any value of $x$ within the closed interval $[-1, 1]$, the sum of the arcsine and arccosine of that value always equals a constant right angle.

2. Mathematical Proof: Let $\sin^{-1} x = \theta$. By the definition of inverse functions, this implies: $$x = \sin \theta$$ Using the co-function identity $\sin \theta = \cos(\frac{\pi}{2} - \theta)$, we can write: $$x = \cos\left(\frac{\pi}{2} - \theta\right)$$ Applying the inverse cosine function to both sides: $$\cos^{-1} x = \frac{\pi}{2} - \theta$$

3. Final Summation: Now, substitute $\theta = \sin^{-1} x$ back into the equation: $$\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$$ $$\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$$ This holds true for all $x \in [-1, 1]$.