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To Find: The value of $\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)$

Let, $x=\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)$

$\Rightarrow \mathrm{x}=\frac{\pi}{3}-\left[\pi-\sec ^{-1}(2)\right]\left(\because \tan \left(\frac{\pi}{3}\right)=\sqrt{3}\right.$ and $\left.\sec ^{-1}(-\theta)=\pi-\sec ^{-1}(\theta)\right)$

$\Rightarrow \mathrm{x}=\frac{\pi}{3}-\left[\pi-\frac{\pi}{3}\right]$

$\Rightarrow \mathrm{x}=\frac{\pi}{3}-\left[\frac{2 \pi}{3}\right]$

$\Rightarrow \mathrm{x}=-\frac{\pi}{3}$