\(VT=\left(\frac{sin^4x-1}{sin^2x}\right)^2+\left(\frac{cos^4x-1}{cos^2x}\right)^2=\left(\frac{\left(sin^2x-1\right)\left(sin^2x+1\right)}{sin^2x}\right)^2+\left(\frac{\left(cos^2x-1\right)\left(cos^2x+1\right)}{cos^2x}\right)^2\)
\(=\left(\frac{cos^2x\left(sin^2x+1\right)}{sin^2x}\right)^2+\left(\frac{sin^2x\left(cos^2x+1\right)}{cos^2x}\right)^2=\left(cos^2x+cot^2x\right)^2+\left(sin^2x+tan^2x\right)^2\)
\(VT\ge\frac{1}{2}\left(cos^2x+cot^2x+sin^2x+tan^2x\right)^2=\frac{1}{2}\left(1+tan^2x+cot^2x\right)^2\)
\(VT\ge\frac{1}{2}\left(1+2\sqrt{tan^2x.cot^2x}\right)^2=\frac{9}{2}\)
\(VP=\frac{9}{2}-\left(siny-1\right)^2\le\frac{9}{2}\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}tan^2x=cot^2x\\siny=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}cos2x=0\\siny=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\y=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
