Q68 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics

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

  • A
    $\dfrac{\pi}{6}$
  • Correct answer
    B
    $\dfrac{\pi}{4}$
  • C
    $\dfrac{\pi}{3}$
  • D
    $\dfrac{\pi}{2}$

Solution

Simplify the expression inside the inverse tangent function:

$\sqrt{5-2\sqrt{6}} = \sqrt{3 + 2 - 2\sqrt{3}\sqrt{2}} = \sqrt{(\sqrt{3} - \sqrt{2})^2} = \sqrt{3} - \sqrt{2}$

Substitute this back into the given equation:

$\tan^{-1}\left[\dfrac{\sqrt{5-2\sqrt{6}}}{1+\sqrt{6}}\right] = \tan^{-1}\left[\dfrac{\sqrt{3}-\sqrt{2}}{1+\sqrt{3}\sqrt{2}}\right]$

Using the identity $\tan^{-1}\left(\dfrac{x-y}{1+xy}\right) = \tan^{-1}(x) - \tan^{-1}(y)$, we get:

$\tan^{-1}\left[\dfrac{\sqrt{3}-\sqrt{2}}{1+\sqrt{3}\sqrt{2}}\right] = \tan^{-1}(\sqrt{3}) - \tan^{-1}(\sqrt{2})$

Since $\tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}$, the expression becomes:

$\dfrac{\pi}{3} - \tan^{-1}(\sqrt{2})$

Equating this to the given right-hand side:

$\dfrac{\pi}{3} - \tan^{-1}(\sqrt{2}) = \dfrac{\pi}{3} - \tan^{-1}(k)$

$\Rightarrow k = \sqrt{2}$

Now, find the value of $\sec^{-1}(k)$:

$\sec^{-1}(\sqrt{2}) = \dfrac{\pi}{4}$

Answer: $\dfrac{\pi}{4}$

Hence, the correct answer is option: 'B'.

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