Question

\(\frac{\sin^2x-2}{\sin^2x-4\cos^2\frac{x}{2}}\)= \(\tan^2\frac{x}{2}\)

Answer 1

ĐKXĐ: ...

\(tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{2sin\frac{x}{2}.cos\frac{x}{2}}{2cos^2\frac{x}{2}}=\frac{sinx}{1+cosx}\)

Thay vào pt:

\(\frac{sin^2x-2}{1-cos^2x-2\left(1+cosx\right)}=\frac{sin^2x}{\left(1+cosx\right)^2}\)

\(\Leftrightarrow\frac{sin^2x-2}{-\left(cos^2x+2cosx+1\right)}=\frac{sin^2x}{\left(1+cosx\right)^2}\)

\(\Leftrightarrow\frac{2-sin^2x}{\left(1+cosx\right)^2}=\frac{sin^2x}{\left(1+cosx\right)^2}\)

\(\Leftrightarrow2-sin^2x=sin^2x\)

\(\Rightarrow sin^2x=1\Rightarrow cosx=0\)

\(\Rightarrow x=\frac{\pi}{2}+k\pi\)