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: Take tangent of both sides:
Let $k$ be unknown. $\tan\left(\frac\pi3-\tan^{-1}k\right)=\dfrac{\sqrt3-k}{1+\sqrt3k}$.

Step 2: Equate:
$\dfrac{\sqrt3-\sqrt2}{1+\sqrt6}=\dfrac{\sqrt3-k}{1+\sqrt3k}$. Trying $k=\sqrt2$ makes the right side $\dfrac{\sqrt3-\sqrt2}{1+\sqrt6}$, an exact match.

Step 3: Evaluate:
$\sec^{-1}\sqrt2=\cos^{-1}\dfrac1{\sqrt2}=\dfrac\pi4$.

Final Answer:
Substituting k = sqrt 2 matches the left side exactly. \[ \boxed{B} \]