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Question-161410




Question Number 161410 by saly last updated on 17/Dec/21
Commented by saly last updated on 17/Dec/21
  Thank you
$$\:\:{Thank}\:{you} \\ $$
Commented by saly last updated on 17/Dec/21
  Do you help me
$$\:\:{Do}\:{you}\:{help}\:{me} \\ $$
Commented by geron last updated on 17/Dec/21
tg(((−27Π)/4))=tg(−6Π−((3Π)/4))=tg(−((3Π)/4))=−((sin(((3Π)/4)))/(cos(((3Π)/4))))=  =−(((√2)/2)/( −((√2)/2)))=1
$${tg}\left(\frac{−\mathrm{27}\Pi}{\mathrm{4}}\right)={tg}\left(−\mathrm{6}\Pi−\frac{\mathrm{3}\Pi}{\mathrm{4}}\right)={tg}\left(−\frac{\mathrm{3}\Pi}{\mathrm{4}}\right)=−\frac{{sin}\left(\frac{\mathrm{3}\Pi}{\mathrm{4}}\right)}{{cos}\left(\frac{\mathrm{3}\Pi}{\mathrm{4}}\right)}= \\ $$$$=−\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\mathrm{1} \\ $$

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