câu trả lời nè:
neu dung may tih' thi sao ma chinh xac duong chu
phai lam the nay,chinh xac dung nanomet ne`,chu may tinh thi noi lam chi
sin a +cos a=7/5 (1)
sin ^2a +cos^2 a=1 (2)
giai he pt (1,2)
(1)=>sin a=7/5-cosa
(2)=>cos^2a+(7/5-cosa)^2=1
giai he pt kia ra (pt bac 2 do',tinh delta)
vi 0<a<90 nen chon nghiem cosa>0
tinh cosa,=>sina=>tga
Ta có : \(tan\alpha=\frac{sin\alpha}{cos\alpha}\); \(sin^2\alpha+cos^2\alpha=1\)
Theo đề bài : \(sin\alpha+cos\alpha=\frac{7}{5}\Leftrightarrow\left(sin\alpha+cos\alpha\right)^2=\frac{49}{25}\Leftrightarrow sin^2\alpha+cos^2\alpha+2sin\alpha.cos\alpha=\frac{49}{25}\)
\(\Leftrightarrow1+2sin\alpha.cos\alpha=\frac{49}{25}\Leftrightarrow sin\alpha.cos\alpha=\frac{\frac{49}{25}-1}{2}=\frac{12}{25}\)
Ta có hệ : \(\hept{\begin{cases}sin\alpha+cos\alpha=\frac{7}{5}\\sin\alpha.cos\alpha=\frac{12}{25}\end{cases}\Leftrightarrow}\hept{\begin{cases}sin\alpha=\frac{4}{5}\\cos\alpha=\frac{3}{5}\end{cases}}\)hoặc \(\hept{\begin{cases}sin\alpha=\frac{3}{5}\\cos\alpha=\frac{4}{5}\end{cases}}\)
Vậy \(tan\alpha=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\) hoặc \(tan\alpha=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}\)
