Question Number 146242 by mathdanisur last updated on 12/Jul/21
$$\mathrm{4}\:{sin}\left(\mathrm{50}°\right)\:-\:\frac{\mathrm{1}}{{cos}\left(\mathrm{20}°\right)}\:=\:? \\ $$
Answered by gsk2684 last updated on 12/Jul/21
$$=\frac{\mathrm{4cos}\:\mathrm{40}^{\mathrm{0}} \mathrm{cos}\:\mathrm{20}^{\mathrm{0}} −\mathrm{1}}{\mathrm{cos}\:\mathrm{20}^{\mathrm{0}} } \\ $$$$=\frac{\mathrm{2}\left(\mathrm{cos}\:\mathrm{60}^{\mathrm{0}} +\mathrm{cos}\:\mathrm{20}^{\mathrm{0}} \right)−\mathrm{1}}{\mathrm{cos}\:\mathrm{20}^{\mathrm{0}} } \\ $$$$=\frac{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{20}^{\mathrm{0}} \right)−\mathrm{1}}{\mathrm{cos}\:\mathrm{20}^{\mathrm{0}} } \\ $$$$=\mathrm{2} \\ $$
Commented by mathdanisur last updated on 12/Jul/21
$${thankyou}\:{Ser} \\ $$