Question

If \(tan^{-1}[\frac{\sqrt{5-2\sqrt{6}}}{1+\sqrt{6}}] = \frac{π}{3}-tan^{-1}(k)\), then \(sec^{-1}(k) = ...\)

• A. \(\frac{π}{6}\)
• B. \(\frac{π}{4}\)
• C. \(\frac{π}{3}\)
• D. \(\frac{π}{2}\)

Hint:

Write 5 - 2 sqrt 6 as (sqrt3 - sqrt2)^2 first.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We use \(\tan^{-1}x-\tan^{-1}y=\tan^{-1}\dfrac{x-y}{1+xy}\) to split the left side.

Step 2: Simplify the root:
\(5-2\sqrt6=(\sqrt3-\sqrt2)^2\), so \(\sqrt{5-2\sqrt6}=\sqrt3-\sqrt2\).

Step 3: Rewrite the denominator:
\(1+\sqrt6=1+\sqrt3\cdot\sqrt2\). So
\[ \tan^{-1}\frac{\sqrt3-\sqrt2}{1+\sqrt3\sqrt2}=\tan^{-1}\sqrt3-\tan^{-1}\sqrt2=\frac\pi3-\tan^{-1}\sqrt2 \]

Step 4: Compare:
So \(k=\sqrt2\) and \(\sec^{-1}\sqrt2=\dfrac\pi4\), because \(\sec\dfrac\pi4=\sqrt2\).

Step 5: Choose:
Option (B).

Final Answer:
k = sqrt(2), so sec inverse of k is pi/4. \[ \boxed{\frac{\pi}{4}} \]