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 \(\sin\alpha=\dfrac{4}{5}\)

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

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

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

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

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

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

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

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

#mã mã#

a) \(\dfrac{2sina+3cosa}{3sina-4cosa}=\dfrac{9}{5}\)

b) \(\dfrac{sina.cosa}{sin^2a-sina.cosa+cos^2a}=0\)

​\(a.\dfrac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-4\cos\alpha}=\dfrac{2\left(3cos\alpha\right)+3cos\alpha}{3\left(3cos\alpha\right)-4cos\alpha}=\dfrac{9cos\alpha}{5cos\alpha}=\dfrac{9}{5}\)
\(b.\dfrac{sin\alpha cos\alpha}{sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{9cos^2\alpha-3cos^2\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{7cos^2\alpha}=\dfrac{3}{7}\)

\(\dfrac{sina+cosa}{sina-cosa}=\dfrac{\dfrac{sina+cosa}{cosa}}{\dfrac{sina-cosa}{cosa}}=\dfrac{tana+1}{tana-1}=\dfrac{3}{1}=3\)

Có \(\dfrac{sin\alpha}{cos\alpha}=tan\alpha=2\)\(\Rightarrow sin\alpha=2cos\alpha\)

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