Question

1+cotx = 1-\(\dfrac{cosx}{sin^2x}\)

Answer 1

cotx=\(\dfrac{\cos x}{\sin x}\left(đk:\sin x\ne0\right)\)

\(\Rightarrow1+\dfrac{\cos x}{\sin x}=1-\dfrac{\cos x}{\sin^2x}\)

\(\Leftrightarrow\dfrac{\cos x}{\sin x}+\dfrac{\cos x}{\sin^2x}=0\)

\(\Leftrightarrow\cos x\sin x+\cos x=0\)

\(\Leftrightarrow\cos x\left(\sin x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos x=0\\\sin x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqcap}{2}+k\sqcap\\x=\dfrac{-\sqcap}{2}+k2\sqcap\end{matrix}\right.\)(tmđk)