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Question Number 189664 by Rupesh123 last updated on 20/Mar/23

Answered by BaliramKumar last updated on 20/Mar/23

tan5°tan(60°−5°)tan(60°+5°)tan75°  tan(3×5°)tan75°  tan15°tan75°  tan15°cot15° = 1

$${tan}\mathrm{5}°{tan}\left(\mathrm{60}°−\mathrm{5}°\right){tan}\left(\mathrm{60}°+\mathrm{5}°\right){tan}\mathrm{75}° \\ $$$${tan}\left(\mathrm{3}×\mathrm{5}°\right){tan}\mathrm{75}° \\ $$$${tan}\mathrm{15}°{tan}\mathrm{75}° \\ $$$${tan}\mathrm{15}°{cot}\mathrm{15}°\:=\:\mathrm{1} \\ $$

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