cotθ=1+cos2θsin2θ=1+cos15osin15o where θ=712∘ =1+3+1223-122=22+3+13-1=22+3+13+13-13+1=26+22+3+3+3+12=2+3+4+6

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$\cot 7 {\frac{1}{2}}^{o} =$

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$\sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4}$

$\sqrt{6} + \sqrt{3} + \sqrt{4} + \sqrt{2}$

$\sqrt{1} + \sqrt{2} + \sqrt{4} + \sqrt{6}$

$\sqrt{3} - \sqrt{2} + \sqrt{6} + \sqrt{4}$

answer is B.

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Detailed Solution

$\cot θ = \frac{1 + \cos 2 θ}{\sin 2 θ} = \frac{1 + \cos 15^{o}}{\sin 15^{o}}$ $(w h e r e θ = 7 {\frac{1}{2}}^{\circ})$

$\begin{array}{l} = \frac{1 + \frac{\sqrt{3} + 1}{2 \sqrt{2}}}{\frac{\sqrt{3} - 1}{2 \sqrt{2}}} \\ = \frac{2 \sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1} \\ = \frac{(2 \sqrt{2} + \sqrt{3} + 1) (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} + 1)} \\ = \frac{2 \sqrt{6} + 2 \sqrt{2} + 3 + \sqrt{3} + \sqrt{3} + 1}{2} \\ = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6} \end{array}$