alleryFor every $x \in [-1, 1]$, the value of $\sin^{-1}(x) + \cos^{-1}(x)$ is
A$2\pi$
B$\tfrac{\pi}{2}$
C$\tfrac{\pi}{4}$
D$\pi$
Answer & Solution
Correct answer: B. $\tfrac{\pi}{2}$
Let $\theta = \sin^{-1}(x)$ so $x = \sin\theta$ with $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$.
Then $\cos\left(\tfrac{\pi}{2} - \theta\right) = \sin\theta = x$, and $\tfrac{\pi}{2}-\theta \in [0,\pi]$ — the range of $\cos^{-1}$.
**So $\cos^{-1}(x) = \tfrac{\pi}{2} - \sin^{-1}(x)$, giving the complementary identity $\sin^{-1}(x) + \cos^{-1}(x) = \tfrac{\pi}{2}$.**
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