Question

tan 1° tan 2° tan 3° ... tan 89° = ?
(a) 1
(b) 0
(c) Cannot be determined
(d) None of these

Answer 1

$$\begin{gathered} (a)\; 1 \\ \text{We have:} \\ {\text{tan1}}^0\tan 2^0\tan 3^0...\tan {89}^0 \\ =\tan 1^0\tan 2^0...\tan {45}^0...\tan {88}^0\tan {89}^0 \\ =\tan 1^0\tan 2^0...\tan {45}^0...\tan {88}^0\tan {89}^0 \\ =\tan 1^0\tan 2^0...\mathrm{t}\mathrm{a}\mathrm{n}{45}^0...\mathrm{t}\mathrm{an}({90}^0-2^0)\tan ({90}^0-1^0) \\ =\tan 1^0\tan 2^0...\cot 2^0\cot 1^0[\because \tan ({90}^0-\mathrm{\theta })=\cot \mathrm{\theta }{\text{and tan45}}^0=1] \\ =\frac{1}{\cot 1^0}\times \frac{1}{\cot 2^0}...1...\cot 2^0\times \cot 1^0 \\ =1 \end{gathered}$$