If tan^{-1} left[ frac{sqrt{5} - 2sqrt{6}}{1 | vedclass

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(B) Given: $	an^{-1} \left[ \frac{\sqrt{5} - 2\sqrt{6}}{1 + \sqrt{6}} ight] = \frac{\pi}{3} - 	an^{-1}(k)$.Note that $\sqrt{5} - 2\sqrt{6} = \sqrt

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$\frac{\pi}{4}$

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(B) Given: $	an^{-1} \left[ \frac{\sqrt{5} - 2\sqrt{6}}{1 + \sqrt{6}} ight] = \frac{\pi}{3} - 	an^{-1}(k)$.Note that $\sqrt{5} - 2\sqrt{6} = \sqrt{3} - \sqrt{2}$ is incorrect; let us simplify the expression inside $	an^{-1}$.Actually, $\frac{\sqrt{3} - \sqrt{2}}{1 + \sqrt{6}} = 	an(	an^{-1}(\sqrt{3}) - 	an^{-1}(\sqrt{2})) = 	an(\frac{\pi}{3} - 	an^{-1}(\sqrt{2}))$.Comparing this with the given equation, we have $	an^{-1}(\sqrt{2}) = 	an^{-1}(k)$, so $k = \sqrt{2}$.We need to find $\sec^{-1}(\sqrt{2})$.Since $\sec(\frac{\pi}{4}) = \sqrt{2}$, it follows that $\sec^{-1}(\sqrt{2}) = \frac{\pi}{4}$.Thus, the correct option is $B$.

If $\tan^{-1} \left[ \frac{\sqrt{5} - 2\sqrt{6}}{1 + \sqrt{6}} \right] = \frac{\pi}{3} - \tan^{-1}(k)$, then $\sec^{-1}(k) = \dots$

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