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a/ Có \(\tan\alpha=\frac{1}{3}\Rightarrow\frac{\sin\alpha}{\cos\alpha}=\frac{1}{3}\Leftrightarrow\cos\alpha=3\sin\alpha\)
Thay vào biểu thức có:
\(\frac{3\sin\alpha+\sin\alpha}{3\sin\alpha-\sin\alpha}=\frac{4\sin\alpha}{2\sin\alpha}=2\)
b/ Có \(\sin\alpha+\cos\alpha=\frac{7}{5}\Rightarrow\sin\alpha=\frac{7}{5}-\cos\alpha\) (1)
Có \(\sin^2\alpha+\cos^2\alpha=1\) (2)
Thay (1) vào (2) rồi tự thay số vào giải PTB2 để tìm cos và sin
Có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
Thay vào là OK
\(M=\frac{\frac{sina}{cosa}+\frac{cosa}{cosa}}{\frac{sina}{cosa}-\frac{cosa}{cosa}}=\frac{tana+1}{tana-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=...\)
\(N=\frac{\frac{sina.cosa}{cos^2a}}{\frac{sin^2a}{cos^2a}-\frac{cos^2a}{cos^2a}}=\frac{tana}{tan^2a-1}=...\) (thay số bấm máy)
\(P=\frac{\frac{sin^3a}{cos^3a}+\frac{cos^3a}{cos^3a}}{\frac{2sina.cos^2a}{cos^3a}+\frac{cosa.sin^2a}{cos^3a}}=\frac{tan^3a+1}{2tana+tan^2a}=...\)
1.\(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\cos^2\alpha=1-\sin^2\alpha=1-\left(\frac{3}{5}\right)^2=1-\frac{9}{25}=\frac{16}{25}\)
\(\Rightarrow\cos\alpha=\frac{4}{5}\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}\)
\(\cot\alpha=\frac{\cos\alpha}{\sin\alpha}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\)
2.\(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\sin^2\alpha=1-\cos^2\alpha=1-\left(0,8\right)^2=1-0,64=0,36\)
\(\Rightarrow\sin\alpha=0,6\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{0,6}{0,8}=\frac{3}{4}\)
\(\tan\alpha.\cot\alpha=1\Rightarrow\cot\alpha=\frac{1}{\tan\alpha}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\)
2. \(\left(\sin a+\cos a\right)^2+\left(\sin a-\cos a\right)^2+2\)
\(=\sin^2a+2.\sin a.\cos a+\cos^2a+\sin^2a\cdot2.\sin a.\cos a+\cos^2a+2\)
\(=2\sin^2a+2\cos^2a+2\)
\(=2\left(\sin^2a+\cos^2a\right)+2\)
\(=2.1+2=4\)
=> biểu thức trên ko phụ thuộc vào a
1. a.) \(\cot a=\dfrac{1}{\tan a}=\dfrac{1}{\sqrt{3}}\)
\(\tan\sqrt{3}=60\Rightarrow a=60^o\)
\(\sin60=\dfrac{\sqrt{3}}{2}\)
\(\cos60=\dfrac{1}{2}\)
b.) \(\cos^2a=1-\left(\dfrac{15}{17}\right)^2=\dfrac{64}{289}\Rightarrow\cos a=\dfrac{8}{17}\)
\(\tan a=\dfrac{\sin a}{\cos a}=\dfrac{\dfrac{15}{17}}{\dfrac{8}{17}}=\dfrac{15}{17}.\dfrac{17}{8}=\dfrac{15}{8}\)
Lời giải:
Ta biết:
$\sin ^2a+\cos ^2a=1$
$\Rightarrow \cos ^2a=1-\sin ^2a=1-(\frac{2}{3})^2=\frac{5}{9}$
$\Rightarrow \cos a=\frac{\sqrt{5}}{3}$
$\tan a=\frac{\sin a}{\cos a}=\frac{2}{3}:\frac{\sqrt{5}}{3}=\frac{2}{\sqrt{5}}$
$\cot a=\frac{1}{\tan a}=\frac{\sqrt{5}}{2}$
Có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha};\cot\alpha=\frac{\cos\alpha}{\sin\alpha}\)
\(\Rightarrow\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha}=8\)
\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha=8\sin\alpha.\cos\alpha\)
\(\Leftrightarrow\sin\alpha.\cos\alpha=\frac{1}{8}\Leftrightarrow2\sin\alpha.\cos\alpha=\frac{1}{4}\)
Có \(\sin^2\alpha+\cos^2\alpha=1\Leftrightarrow\sin^2\alpha+2\sin\alpha.\cos\alpha+\cos^2\alpha=1+2\sin\alpha.\cos\alpha\)
\(\Leftrightarrow\left(\sin\alpha+\cos\alpha\right)^2=1+\frac{1}{4}=\frac{5}{4}\)
\(\Leftrightarrow\sin\alpha+\cos\alpha=\frac{\sqrt{5}}{2}\)
a)\(\sin\alpha=\dfrac{9}{15}\Rightarrow\sin^2\alpha=\dfrac{81}{225}\)
Có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\cos^2\alpha=1-\sin^2\alpha=1-\dfrac{81}{225}=\dfrac{144}{225}\)
\(\Rightarrow\cos\alpha=\sqrt{\dfrac{144}{225}}=\dfrac{12}{15}=\dfrac{4}{5}\)
\(\Rightarrow\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{9}{15}:\dfrac{4}{5}=\dfrac{3}{4}\)
\(\cot\alpha=\dfrac{\cos\alpha}{\tan\alpha}=\dfrac{4}{5}:\dfrac{9}{15}=\dfrac{4}{3}\)
b)\(\cos\alpha=\dfrac{3}{5}\Rightarrow\cos^2\alpha=\dfrac{9}{25}\)
Có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\sin^2\alpha=1-\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\)
\(\Rightarrow\sin\alpha=\dfrac{4}{5}\)
\(\Rightarrow\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)
\(\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\)
ủa cos a=3 đc hả