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Question Number 201460 by hardmath last updated on 06/Dec/23
Find the smallest positive period of the  function:  y = ∣ tan 2x ∣  +  ∣ cot 2x ∣
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{positive}\:\mathrm{period}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{function}: \\ $$$$\mathrm{y}\:=\:\mid\:\mathrm{tan}\:\mathrm{2x}\:\mid\:\:+\:\:\mid\:\mathrm{cot}\:\mathrm{2x}\:\mid \\ $$
Answered by Mathspace last updated on 07/Dec/23
y(x)=∣tan(2x)∣+(1/(∣tan(2x)∣))  y(x+(π/2))=∣tan(2(x+(π/2)))+(1/(∣tan(2(x+(π/2)))))  =∣tan(2x+π)∣+(1/(∣tan(2x+π)∣))  =∣tan(2x)∣+(1/(∣tan(2x)∣))  =f(x) so the small period is  T=(π/2)
$${y}\left({x}\right)=\mid{tan}\left(\mathrm{2}{x}\right)\mid+\frac{\mathrm{1}}{\mid{tan}\left(\mathrm{2}{x}\right)\mid} \\ $$$${y}\left({x}+\frac{\pi}{\mathrm{2}}\right)=\mid{tan}\left(\mathrm{2}\left({x}+\frac{\pi}{\mathrm{2}}\right)\right)+\frac{\mathrm{1}}{\mid{tan}\left(\mathrm{2}\left({x}+\frac{\pi}{\mathrm{2}}\right)\right)} \\ $$$$=\mid{tan}\left(\mathrm{2}{x}+\pi\right)\mid+\frac{\mathrm{1}}{\mid{tan}\left(\mathrm{2}{x}+\pi\right)\mid} \\ $$$$=\mid{tan}\left(\mathrm{2}{x}\right)\mid+\frac{\mathrm{1}}{\mid{tan}\left(\mathrm{2}{x}\right)\mid} \\ $$$$={f}\left({x}\right)\:{so}\:{the}\:{small}\:{period}\:{is} \\ $$$${T}=\frac{\pi}{\mathrm{2}} \\ $$

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