Problem

\(tan \theta = 2k\)

\(tan \frac{\theta}{2} = \sqrt {\frac {1-cos \theta}{1+ cos \theta}}\)

\(tan \frac{\theta}{2} = \sqrt{\frac{1-\frac{1}{\sqrt{4k^2 +1}}}{1+\frac{1}{\sqrt{4k^2 +1}}}}\)

\(tan \frac{\theta}{2} = \sqrt{\frac{\sqrt{4k^2 +1}-1}{\sqrt{4k^2 1}+1}}\)

\(tan \frac{\theta}{2} = \frac{2k}{\sqrt{4k^2 +1}+1}\)

Thus, \(b = (k \sqrt{k^2 + \frac{1}{4}}) (\frac{2k}{\sqrt{4k^2 +1}+1})\)

\(= \frac{k^2 \sqrt{4k^2 +1}}{\sqrt{4k^2 +1}+1}\)

Thus, \(a = \frac{k}{2} (\sqrt{4k^2 +1}+1)\)

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