Question

If \(\tan ^{-1}\left[\frac{\sqrt{5+2\sqrt{6}}}{1-\sqrt{6}}\right]=a+\tan ^{-1}(\sqrt{2})\), then \(\csc (a)=\ldots\)

• A. \(2\)
• B. \(-2\)
• C. \(\frac{2}{\sqrt{3}}\)
• D. \(\frac{-2}{\sqrt{3}}\)

Hint:

Simplify the surd, use the tan(A - B) formula to get tan a, then fix the quadrant from the principal value range of the inverse tangent.

Answer 1

The Correct Option is D

Step 1: Simplify the surd
\(5+2\sqrt{6}=(\sqrt{2}+\sqrt{3})^{2}\), so the argument is \(T=\frac{\sqrt{2}+\sqrt{3}}{1-\sqrt{6}}\).

Step 2: Find tan a
\(a=\tan ^{-1}T-\tan ^{-1}\sqrt{2}\), so \(\tan a=\frac{T-\sqrt{2}}{1+\sqrt{2}T}\).
Numerator: \(\frac{3\sqrt{3}}{1-\sqrt{6}}\). Denominator: \(\frac{3}{1-\sqrt{6}}\). So \(\tan a=\sqrt{3}\).

Step 3: Choose the correct angle
\(T\approx -2.17\), so \(\tan ^{-1}T\approx -1.14\) and \(\tan ^{-1}\sqrt{2}\approx 0.955\).
Thus \(a\approx -2.09=-\frac{2\pi }{3}\), not \(\frac{\pi }{3}\).

Step 4: Find csc a
\(\csc \left(-\frac{2\pi }{3}\right)=\frac{1}{-\sin \frac{2\pi }{3}}=-\frac{2}{\sqrt{3}}\).

Final Answer:
The value of \(\csc a\) is \(-\frac{2}{\sqrt{3}}\). \[ \boxed{\text{(D) }\dfrac{-2}{\sqrt{3}}} \]