ko bt làm xuống lớp 8 đê
\(tana\cdot cota=1\)
\(tana\cdot\frac{2}{3}=1\)
\(tana=\frac{3}{2}\)
\(1+tan^2a=\frac{1}{cos^2a}\)
\(1+\left(\frac{3}{2}\right)^2=\frac{1}{cos^2a}\)
\(1+\frac{9}{4}=\frac{1}{cos^2a}\)
\(\frac{13}{4}=\frac{1}{cos^2a}\)
\(cos^2a=\frac{4}{13}\)
\(cosa=\frac{2\sqrt{13}}{13}\) ( cấp 2 nên chỉ lấy cos dương )
\(sin^2a+cos^2a=1\)
\(sin^2a+\frac{4}{13}=1\)
\(sin^2a=\frac{9}{13}\)
\(sin^2a+cos^3a-tana\)
\(=\frac{9}{13}+\frac{4\sqrt{13}}{13}-\frac{3}{2}\)
\(=\frac{18}{26}+\frac{8\sqrt{13}}{26}-\frac{39}{26}\)
\(=\frac{-21+8\sqrt{13}}{26}\)
