giup to lam bai nay di 🪷️🪻

$\text{Đáp án+chi tiết:}$

`b)` `TXĐ: D=R\\ {pi/2+kpi;(-pi)/2+k2pi|kinZZ}`

`sin^2x+cos^2x=1=>cos^2x=1-sin^2x` (1)$\\$$\text{Ta có: VT}$`=1/cosx-tanx`$\\$ ㅤㅤ             ㅤ`=1/cosx-sinx/cosx`$\\$ ㅤㅤ             ㅤ`=(1-sinx)/cosx`$\\$ ㅤㅤ             ㅤ`=((1-sinx).(1+sinx))/(cosx.(1+sinx))`$\\$ ㅤㅤ             ㅤ`=(1-sin^2x)/(cosx.(1+sinx))`$\\$ ㅤㅤ             ㅤ`=cos^2x/(cosx.(1+sinx))`ㅤㅤ(Thay (1) vào)$\\$ ㅤㅤ             ㅤ`=cosx/(1+sinx)=VP`ㅤㅤ`(đpcm)`$\\$ `Vậy`  `1/cosx-tanx=cosx/(1+sinx)`