Prove sin⁻¹x + cos⁻¹x = π/2

Question

Prove that $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ for all $x \in [-1, 1]$.

This is a standard CBSE 12 proof question — it appeared directly in the 2024 Board Exam. Two marks, fully doable if you know the trick.

Solution — Step by Step

Why This Works

The real reason behind this identity is the geometry of the unit circle. If $\sin^{-1}x = \theta$, that angle $\theta$ is measured from the positive x-axis. The angle that gives the same $x$ as a cosine value is exactly the complementary angle — they always add up to $90°$.

Think of a right triangle with hypotenuse 1 and one side $x$. If one acute angle is $\theta$ (opposite side = $x$, so $\sin\theta = x$), then the other acute angle is $\dfrac{\pi}{2} - \theta$ (adjacent side = $x$, so $\cos\left(\dfrac{\pi}{2} - \theta\right) = x$). The two angles always sum to $\dfrac{\pi}{2}$.

The domain check in Step 3 is what makes the proof rigorous. Without checking that $\dfrac{\pi}{2} - \theta$ falls inside $[0, \pi]$, you haven’t technically justified the step where you apply $\cos^{-1}$.

Alternative Method

We can also prove this by defining $f(x) = \sin^{-1}x + \cos^{-1}x$ and showing it’s constant.

Differentiate: $f'(x) = \dfrac{1}{\sqrt{1-x^2}} + \left(-\dfrac{1}{\sqrt{1-x^2}}\right) = 0$

Since $f'(x) = 0$ for all $x \in (-1, 1)$, $f(x)$ is constant on $[-1, 1]$.

Evaluate at $x = 0$: $f(0) = \sin^{-1}(0) + \cos^{-1}(0) = 0 + \dfrac{\pi}{2} = \dfrac{\pi}{2}$.

Therefore $f(x) = \dfrac{\pi}{2}$ for all $x \in [-1, 1]$.

Common Mistake