NIMCET 2026 — Mathematics PYQ

To solve this problem, we must apply the properties of inverse trigonometric functions, specifically focusing on their principal value branches.

1. Evaluating the first term: ${\cos }^{-1}(\cos (-\frac{\pi }{6}))$

Since $\cos (-x)=\cos (x)$, we have $\cos (-\frac{\pi }{6})=\cos (\frac{\pi }{6})$.

The principal value range for ${\cos }^{-1}(x)$ is $[0,\pi ]$. Since $\frac{\pi }{6}$ lies within this range:

$${\cos }^{-1}(\cos (-\frac{\pi }{6}))=\frac{\pi }{6}$$

2. Evaluating the second term: ${\sin }^{-1}(\sin (\frac{5\pi }{6}))$

The principal value range for ${\sin }^{-1}(x)$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$. The angle $\frac{5\pi }{6}$ is outside this range.

We use the identity $\sin (\pi -\theta )=\sin (\theta )$:

$$\sin \left(\frac{5\pi }{6}\right)=\sin \left(\pi -\frac{\pi }{6}\right)=\sin \left(\frac{\pi }{6}\right)$$

Since $\frac{\pi }{6}$ is within the principal range:

$${\sin }^{-1}(\sin (\frac{5\pi }{6}))={\sin }^{-1}(\sin (\frac{\pi }{6}))=\frac{\pi }{6}$$

3. Calculating the total sum:

$$\frac{\pi }{6}+\frac{\pi }{6}=\frac{2\pi }{6}=\frac{\pi }{3}$$

Correct Option: (b)