1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)

\(C=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha\)

\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)

\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)

1: \(cota=\sqrt{5}\)

=>\(cosa=\sqrt{5}\cdot sina\)

\(1+cot^2a=\dfrac{1}{sin^2a}\)

=>\(\dfrac{1}{sin^2a}=1+5=6\)

=>\(sin^2a=\dfrac{1}{6}\)

\(C=sin^2a-sina\cdot\sqrt{5}\cdot sina+\left(\sqrt{5}\cdot sina\right)^2\)

\(=sin^2a\left(1-\sqrt{5}+5\right)=\dfrac{1}{6}\cdot\left(6-\sqrt{5}\right)\)

2: tan a=3

=>sin a=3*cosa 

\(1+tan^2a=\dfrac{1}{cos^2a}\)

=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)

\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)

\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)

\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)

Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)

\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)

$\begin{cases}sinα+cosα=\dfrac{7}{5}\\sin^2α+cos^2α=1\\\end{cases}$

`<=>` $\begin{cases}sinα+cosα=\dfrac{7}{5}\\(sinα+cosα)^2-2sinαcosα=1\\\end{cases}$

`<=>` $\begin{cases}sinα+cosα=\dfrac{7}{5}\\sinα.cosα=\dfrac{12}{25}\\\end{cases}$

`<=>` \(\left\{{}\begin{matrix}\left[{}\begin{matrix}sinα=\dfrac{4}{5}\\cosα=\dfrac{3}{5}\end{matrix}\right.\\\left[{}\begin{matrix}sinα=\dfrac{3}{5}\\cosα=\dfrac{4}{5}\end{matrix}\right.\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}tanα=\dfrac{3}{4}\\tanα=\dfrac{4}{3}\end{matrix}\right.\)

Vậy...

Ta có: \(\left(\sin\alpha+\cos\alpha\right)^2=\dfrac{49}{25}\)

\(\Leftrightarrow2\cdot\sin\alpha\cdot\cos\alpha=\dfrac{49}{25}-1=\dfrac{24}{25}\)

Ta có: \(\left(\sin\alpha-\cos\alpha\right)^2\)

\(=\sin^2\alpha+\cos^2\alpha-\dfrac{24}{25}\)

\(=1-\dfrac{24}{25}=\dfrac{1}{25}\)

\(\Leftrightarrow\sin\alpha-\cos\alpha=\dfrac{1}{5}\)

mà \(\sin\alpha+\cos\alpha=\dfrac{7}{5}\)

nên \(2\cdot\sin\alpha=\dfrac{8}{5}\)

hay \(\sin\alpha=\dfrac{4}{5}\)

\(\Leftrightarrow\cos\alpha=\dfrac{7}{5}-\dfrac{4}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)

\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)

\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)

\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)

\(\dfrac{1}{cos^2\alpha}=1+tan^2\alpha=1+\left(\dfrac{7}{24}\right)^2=\dfrac{625}{576}\)

\(\Rightarrow cos^2\alpha=\dfrac{576}{625}\)

\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{24}{7}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow cos^2\alpha=\dfrac{576}{625}\Rightarrow cos\alpha=\dfrac{24}{25}\)

\(1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\Rightarrow sin^2\alpha=\dfrac{49}{625}\Rightarrow cos\alpha=\dfrac{7}{25}\)

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 \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\)= \(\frac{3\sin\alpha+\sin\alpha}{3\sin\alpha-\sin\alpha}\)= \(\frac{4\sin\alpha}{2\sin\alpha}\)= \(2\)

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