Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, ta có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
\(\frac{1}{3}\)= \(\frac{\sin\alpha}{\cos\alpha}\)
\(\cos\alpha\)= 3 \(\sin\alpha\)
ta có \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\)= \(\frac{3\sin\alpha+\sin\alpha}{3\sin\alpha-\sin\alpha}\)= \(\frac{4\sin\alpha}{2\sin\alpha}\)= \(2\)
#mã mã#
Ta có \(tan^2\alpha+1=\frac{1}{cos^2\alpha}\Rightarrow\frac{1}{cos^2\alpha}=5\Rightarrow cos^2\alpha=\frac{1}{5}\)
Do \(tan\alpha=2\) nên \(cos\alpha\ne0\Rightarrow\frac{A}{cos^2\alpha}=\frac{sin^2\alpha+sin\alpha cos\alpha-3cos^2\alpha}{cos^2\alpha}=tan^2\alpha+tan\alpha-3=3\)
Vậy \(A=3.\frac{1}{5}=\frac{3}{5}\)
\(sin^6a+cos^6a+3sin^2a.cos^2a=sin^6a+cos^6a+3sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)
\(=\left(sin^2a+cos^2a\right)^3=1\)
a, ta có \(\cos^2\alpha\)+ \(\sin^2\alpha\)= 1
1/5 + \(\cos^2\alpha\)= 1
\(\cos^2\alpha\)= 4/5
\(4\cos^2\alpha\)+6 \(\sin^2\alpha\)= 4 . 4/5 + 6.1/5=22/5
b, \(\sin\alpha\)= 2/3
\(\sin^2\alpha\)= 4/9
\(\cos^2\alpha=\frac{5}{9}\)
\(5\cos^2\alpha+2\sin^2=\frac{5.5}{9}+\frac{2.4}{9}=\frac{33}{9}\)
#mã mã#