Question:

Given: Expression $\tan^{-1}(-\sqrt{3})$.
Step 1: Recall Principal Value Branch of $\tan^{-1} x$:
The principal value branch of $\tan^{-1} x$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. [0.5 Mark]
Step 2: Apply Inverse Tangent Property:
Using identity $\tan^{-1}(-x) = -\tan^{-1}(x)$ for all $x \in \mathbb{R}$:
$$\tan^{-1}(-\sqrt{3}) = -\tan^{-1}(\sqrt{3})$$ [0.5 Mark]
Step 3: Evaluate Angle:
Since $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$, we have $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.
$$\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$$ [1.0 Mark]
Conclusion: The principal value of $\tan^{-1}(-\sqrt{3})$ is $-\dfrac{\pi}{3}$.

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🎯 Official CBSE Marking Scheme:
Stating principal value branch (-pi/2, pi/2): 0.5 Mark
Applying identity tan^-1(-x) = -tan^-1(x): 0.5 Mark
Evaluating final angle -pi/3: 1.0 Mark