Question

cot 15° cot 16° cot 17° ... cot 73° cot 74° cot 75° = ?
(a) $\frac{1}{2}$
(b) 0
(c) 1
(d) −1

Answer 1

$$\begin{gathered} (c)\; 1 \\ \text{We have:} \\ {\text{cot15}}^0\cot {16}^0\cot {17}^0...\cot {73}^0\cot {74}^0\cot {75}^0 \\ ={\text{cot15}}^0.\cot {16}^0.\cot {17}^0...\cot {45}^0...\cot {73}^0.\cot {74}^0.\cot {75}^0 \\ ={\text{cot15}}^0.\cot {16}^0.\cot {17}^0...\cot {45}^0...\cot ({90}^0-{17}^0).\cot ({90}^0-{16}^0).\cot ({90}^0-{15}^0) \\ ={\text{cot15}}^0.\cot {16}^0.\cot 17...1...\tan {17}^0.\tan {16}^0.\tan {15}^0[\because \cot ({90}^0-\theta )=\tan \theta {\text{and cot45}}^0=1] \\ =1 \end{gathered}$$