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| 1. |
Find the value of tan1°tan2°tan3........tan89°. |
| Answer» tan1°tan2°tan3°.........tan89°=tan(90°-89°)tan(90°-88°)tan(90°-87°)...............tan(90°-46°)tan45°tan46°.....tan89° = cot89°cot88°cot87°..........cot46°tan45°tan46°......tan89° [by using identity tan(90°-A)= cotA] ={tex}1\\over tan89°{/tex}×{tex}1\\over tan88°{/tex}×.........{tex}1\\over tan46°{/tex}×tan45°×tan46°×......tan88°×tan89° = tan45° =1 | |