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\(A=4\left(1-sin^2x\right)-6sin^2a=4-10sin^2a=4-10.\left(\frac{1}{5}\right)^2=...\)
\(tana+cota=3\Leftrightarrow\frac{sina}{cosa}+\frac{cosa}{sina}=3\Leftrightarrow\frac{sin^2a+cos^2a}{sina.cosa}=3\)
\(\Leftrightarrow\frac{1}{sina.cosa}=3\Leftrightarrow sina.cosa=\frac{1}{3}\)
\(C=cot^2a-cos^2a.cot^2a=cot^2a\left(1-cos^2a\right)=cot^2a.sin^2a\)
\(=\frac{cos^2a}{sin^2a}.sin^2a=cos^2a=1-sin^2a=1-\left(\frac{3}{4}\right)^2=...\)
a.Ta có \(\tan\alpha.\cot\alpha=1\Rightarrow\tan\alpha=\frac{1}{\cot\alpha}\)
\(\Rightarrow\frac{1}{\cot\alpha}+\cot\alpha=2\Rightarrow\cot^2\alpha-2\cot\alpha+1=0\)
\(\cot\alpha=1\Rightarrow\alpha=45^0\)
b.Ta có \(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos^2\alpha=1-\sin^2\alpha\)
\(\Rightarrow7.\sin^2\alpha+5\left(1-\sin^2\alpha\right)=\frac{13}{2}\)\(\Leftrightarrow\sin^2\alpha=\frac{3}{4}\Leftrightarrow\orbr{\begin{cases}sin\alpha=\frac{\sqrt{3}}{2}\\sin\alpha=\frac{-\sqrt{3}}{2}\end{cases}}\)
\(\Rightarrow\alpha=60^0\)
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: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\sin^2\alpha=1-\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\)
\(\Rightarrow\sin\alpha=\dfrac{4}{5}\)
\(\Rightarrow\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)
\(\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\)
Ta có:
\(\hept{\begin{cases}cosa-sina=\frac{1}{5}\\sin^2a+cos^2a=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}cosa=\frac{1}{5}+sina\left(1\right)\\sin^2a+\left(\frac{1}{5}+sina\right)^2=1\left(2\right)\end{cases}}\)
\(\Rightarrow\left(2\right)\Leftrightarrow25sin^2a+5sina-12=0\)
\(\Leftrightarrow\orbr{\begin{cases}sina=-\frac{4}{5}\left(l\right)\\sina=\frac{3}{5}\end{cases}}\)
\(\Rightarrow cosa=\frac{4}{5}\)
\(\Rightarrow\hept{\begin{cases}tana=\frac{3}{4}\\cota=\frac{4}{3}\end{cases}}\)
Gấp gáp chi em cuộc sống vẫn rực rỡ sắc màu
Chim vẫn reo ca và môi hôn đang đứng đợi
Hoa vẫn nở và xuân thì đương tới
Hãy trải lòng xao xuyến với tình yêu.
\(1+tan^2a=1+\frac{sin^2a}{cos^2a}=\frac{cos^2a+sin^2a}{cos^2a}=\frac{1}{cos^2a}\)
\(1+cot^2a=1+\frac{cos^2a}{sin^2a}=\frac{sin^2a+cos^2a}{sin^2a}=\frac{1}{sin^2a}\)
\(cot^2a-cos^2a=\frac{cos^2a}{sin^2a}-cos^2a=cos^2a\left(\frac{1}{sin^2a}-1\right)=cos^2a\left(\frac{1-sin^2a}{sin^2a}\right)\)
\(=cos^2a\left(\frac{cos^2a}{sin^2a}\right)=cos^2a.cot^2a\)
\(\frac{1+cosa}{sina}=\frac{sina\left(1+cosa\right)}{sin^2a}=\frac{sina\left(1+cosa\right)}{1-cos^2a}=\frac{sina\left(1+cosa\right)}{\left(1-cosa\right)\left(1+cosa\right)}=\frac{sina}{1-cosa}\)
\(\sin\alpha=\frac{5}{13}\) => \(\sin^2\alpha=\frac{25}{169}\)
Mà \(\cos^2\alpha+\sin^2\alpha=1\) nên \(\cos^2\alpha=\frac{144}{169}\) => \(\cos\alpha=\frac{12}{13}\)
Ta có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\) => \(\tan\alpha=\frac{5}{13}:\frac{12}{13}=\frac{5}{12}\)
Lại có \(\cot\alpha=\frac{\cos\alpha}{\sin\alpha}\) => \(\cot\alpha=\frac{12}{13}:\frac{5}{13}=\frac{12}{5}\)
Chúc bạn làm bài tốt
Ta có \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow\cos^2\alpha=1-\sin^2\alpha\)
\(\Rightarrow\cos^2\alpha=1-\frac{25}{169}=\frac{144}{169}\Rightarrow\cos\alpha=\frac{12}{13}\)
Ta lại có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{5}{13}.\frac{13}{12}=\frac{5}{12}\Rightarrow\cot\alpha=\frac{12}{5}\)