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Question Number 148773 by mim24 last updated on 31/Jul/21

Answered by som(math1967) last updated on 31/Jul/21

(1/(sin20)) +((√3)/(cos20))  =((cos20+(√3)sin20)/(sin20cos20))  =((2((1/2)cos20+((√3)/2)sin20))/(sin20cos20))  =((2.2(cos60cos20+sin60sin20))/(2sin20cos20))  =((4cos(60−20))/(sin40))  =((4cos40)/(sin40))=4cot40=4tan(90−40)=4tan50

$$\frac{\mathrm{1}}{\boldsymbol{{sin}}\mathrm{20}}\:+\frac{\sqrt{\mathrm{3}}}{\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\boldsymbol{{cos}}\mathrm{20}+\sqrt{\mathrm{3}}\boldsymbol{{sin}}\mathrm{20}}{\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{cos}}\mathrm{20}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\boldsymbol{{sin}}\mathrm{20}\right)}{\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}.\mathrm{2}\left(\boldsymbol{{cos}}\mathrm{60}\boldsymbol{{cos}}\mathrm{20}+\boldsymbol{{sin}}\mathrm{60}\boldsymbol{{sin}}\mathrm{20}\right)}{\mathrm{2}\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{4}\boldsymbol{{cos}}\left(\mathrm{60}−\mathrm{20}\right)}{\boldsymbol{{sin}}\mathrm{40}} \\ $$$$=\frac{\mathrm{4}\boldsymbol{{cos}}\mathrm{40}}{\boldsymbol{{sin}}\mathrm{40}}=\mathrm{4}\boldsymbol{{cot}}\mathrm{40}=\mathrm{4}\boldsymbol{{tan}}\left(\mathrm{90}−\mathrm{40}\right)=\mathrm{4}\boldsymbol{{tan}}\mathrm{50} \\ $$

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