Đề là \(tanx=cotx+\frac{1}{cosx}\) hay \(tanx=\frac{cotx+1}{cosx}\) bạn?
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\frac{sinx}{cosx}=\frac{cosx}{sinx}+\frac{1}{cosx}\)
\(\Leftrightarrow sin^2x=cos^2x+sinx\)
\(\Leftrightarrow sin^2x=1-sin^2x+sinx\)
\(\Leftrightarrow2sin^2x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sinx+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\left(l\right)\\x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)