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Lời giải:
Ta có:
$\sin ^2a=1-\cos ^2a=1-(\frac{3}{5})^2=\frac{16}{25}$
$0< a< 90$ nên $\sin a>0$. Do đó $\sin a=\frac{4}{5}$
$\tan a=\frac{\sin a}{\cos a}=\frac{4}{5}: \frac{3}{5}=\frac{4}{3}$
$\cot a=\frac{1}{\tan a}=\frac{3}{4}$
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#
đặt \(\sin\alpha=a;\cos\alpha=b\)
khi đó:
\(a+b=\frac{7}{5}\Leftrightarrow a^2+b^2+2ab=\frac{49}{25}\)
\(\Leftrightarrow1+2ab=\frac{49}{25}\Leftrightarrow2ab=\frac{24}{25}\Leftrightarrow ab=\frac{12}{25}\)
ta có
\(\left\{{}\begin{matrix}a+b=\frac{7}{5}\\ab=\frac{12}{25}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{7}{5}-b\\\left(\frac{7}{5}-b\right)b=\frac{12}{25}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{7}{5}-b\\b^2-\frac{7}{5}b+\frac{12}{25}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{7}{5}-b\\\left(b-\frac{3}{5}\right)\left(b-\frac{4}{5}\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{7}{5}-b\\\left[{}\begin{matrix}b=\frac{3}{5}\\b=\frac{4}{5}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\frac{3}{5}\\b=\frac{4}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\frac{4}{5}\\b=\frac{3}{5}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{a}{b}=\frac{3}{4}\\\frac{a}{b}=\frac{4}{3}\end{matrix}\right.\)\(\)
hay tan \(\alpha\approx37^o\)hoặc tan\(\alpha\approx53^o\)
đời cha tao đã đi làm xe ôm
1 + tan2a = 1 + \(\frac{sin^2a}{cos^2a}\)=\(\frac{cos^2a}{cos^2a}+\frac{sin^2a}{cos^2a}=\frac{cos^2a+sin^2a}{cos^2a}=\frac{1}{cos^2a}\)