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MHTCET 15 Apr 2026 Shift 1 Paper

Question : 117 of 150

Marks: +1, -0

\tan ^{-1}\left[\frac{\sqrt{5-2 \sqrt{6}}}{1+\sqrt{6}}\right]=\frac{\pi}{3}-\tan ^{-1}(k)

\sec ^{-1}(k)=\ldots

Solution:

Concept:

Simplify the nested radical and use the tangent difference formula to find

k

, then evaluate

\sec^{-1}(k)

Explanation:

Given:

\tan^{-1}\left[\frac{\sqrt{5-2\sqrt{6}}}{1+\sqrt{6}}\right] = \frac{\pi}{3} - \tan^{-1}(k)

First simplify

5-2\sqrt{6}

(\sqrt{3}-\sqrt{2})^2

Hence,

\sqrt{5-2\sqrt{6}} = \sqrt{3}-\sqrt{2}

So the equation becomes:

\tan^{-1}\left[\frac{\sqrt{3}-\sqrt{2}}{1+\sqrt{6}}\right] = \frac{\pi}{3} - \tan^{-1}(k)

Rearrange:

\tan^{-1}(k) = \frac{\pi}{3} - \tan^{-1}\left[\frac{\sqrt{3}-\sqrt{2}}{1+\sqrt{6}}\right]

Take tangent on both sides:

k = \tan\left(\frac{\pi}{3} - \tan^{-1}\left[\frac{\sqrt{3}-\sqrt{2}}{1+\sqrt{6}}\right]\right)

Using

\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

k = \frac{\sqrt{3} - \frac{\sqrt{3}-\sqrt{2}}{1+\sqrt{6}}}{1 + \sqrt{3} \cdot \frac{\sqrt{3}-\sqrt{2}}{1+\sqrt{6}}}

After simplification, we get:

k = \sqrt{2}

Now compute:

\sec^{-1}(k) = \sec^{-1}(\sqrt{2})

Since

\sec\frac{\pi}{4} = \sqrt{2}

, we have:

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

Answer:

\frac{\pi}{4}

Option B.

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