${\tan }^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+{\sec }^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)$ is equal to [JEE Main 2023, 24 Jan (Shift 1)]

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\({\tan }^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+{\sec }^{-1}\left(\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right)…

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