$\cot 16 ° \cot 44 ° + \cot 44 ° \cot 76 ° - \cot 76 ° \cot 16 ° =$

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answer is C.

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Detailed Solution

$\cot (A + B) = \frac{\cot A \cdot \cot B - 1}{\cot A + \cot B} \Rightarrow \begin{array}{l} \cot (60 °) = \cot (16 ° + 44 °) \\ \frac{1}{\sqrt{3}} = \frac{\cot 16 ° \cot 44 ° - 1}{\cot 16 ° + \cot 44 °} \\ \cot 16 ° \cot 44 ° = \frac{\cot 16 ° + \cot 44 °}{\sqrt{3}} - 1 \dots \dots . (1) \\ \begin{array}{l} s i m i l l a r l y \cot 44 ° \cot 76 ° = \frac{\cot 44 ° + \cot 76 °}{- \sqrt{3}} + 1 \dots \dots (2) \\ \cot 76 ° \cot 16 ° = \frac{\cot 16 ° - \cot 76 °}{\sqrt{3}} - 1 \dots \dots \dots (3) \\ \begin{array}{l} \cot 16 \cot 44 ° + \cot 44 ° \cot 76 ° - \cot 76 ° \cot 16 ° \\ = \frac{\cot 16 ° + \cot 44}{\sqrt{3}} + 1 - [\frac{\cot 44 ° + \cot 76 °}{\sqrt{3}}] + 1 - [\frac{\cot 16 ° - \cot 76 °}{\sqrt{3}}] + 1 \\ = 3 \end{array} \end{array} \end{array}$