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

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

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 

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

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

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

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

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

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

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

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

a)

\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)

\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)

\(=2\sin ^2a\)

b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)

\(=1+\cos ^2a-1=\cos ^2a\)

\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)

c)

\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)

\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)

\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)

d)

\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)

\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)

f)

\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)

\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)

\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)

a: \(VT=\dfrac{\left(sina+cosa\right)^3-3\cdot sina\cdot cosa\left(sina+cosa\right)}{sina+cosa}\)

=(sina+cosa)^2-3*sina*cosa

=sin^2a+cos^2a-sina*cosa

=1-sina*cosa=VP

c: VT=(sin^2a+cos^2a)^2-2*sin^2a*cos^2a-(sin^2a+cos^2a)^3+3*sin^2a*cos^2a*(sin^2a+cos^2a)

=1-2sin^2a*cos^2a-1+3*sin^2a*cos^2a

=sin^2a*cos^2a=VP

a) Ta có $A=\genfrac{}{}{0ex}{}{\tan \alpha +3\genfrac{}{}{0ex}{}{1}{\tan \alpha }}{\tan \alpha +\genfrac{}{}{0ex}{}{1}{\tan \alpha }}=\genfrac{}{}{0ex}{}{{\tan }^2\alpha +3}{{\tan }^2\alpha +1}=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{1}{{\cos }^2\alpha }+2}{\genfrac{}{}{0ex}{}{1}{{\cos }^2\alpha }}=1+2{\cos }^2\alpha$ Suy ra $A=1+2\cdot \genfrac{}{}{0ex}{}{9}{16}=\genfrac{}{}{0ex}{}{17}{8}$.

b) $B=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{\sin \alpha }{{\cos }^3\alpha }-\genfrac{}{}{0ex}{}{\cos \alpha }{{\cos }^3\alpha }}{\genfrac{}{}{0ex}{}{{\sin }^3\alpha }{{\cos }^3\alpha }+\genfrac{}{}{0ex}{}{3{\cos }^3\alpha }{{\cos }^3\alpha }+\genfrac{}{}{0ex}{}{2\sin \alpha }{{\cos }^3\alpha }}=\genfrac{}{}{0ex}{}{\tan \alpha \left({\tan }^2\alpha +1\right)-\left({\tan }^2\alpha +1\right)}{{\tan }^3\alpha +3+2\tan \alpha \left({\tan }^2\alpha +1\right)}$.

Suy ra $B=\genfrac{}{}{0ex}{}{\sqrt{2}(2+1)-(2+1)}{2\sqrt{2}+3+2\sqrt{2}(2+1)}=\genfrac{}{}{0ex}{}{3(\sqrt{2}-1)}{3+8\sqrt{2}}$.

Lời giải:
a.

$\tan a+\cot a=2\Leftrightarrow \tan a+\frac{1}{\tan a}=2$

$\Leftrightarrow \frac{\tan ^2a+1}{\tan a}=2$

$\Leftrightarrow \tan ^2a-2\tan a+1=0$

$\Leftrightarrow (\tan a-1)^2=0\Rightarrow \tan a=1$

$\cot a=\frac{1}{\tan a}=1$

$1=\tan a=\frac{\cos a}{\sin a}\Rightarrow \cos a=\sin a$

Mà $\cos ^2a+\sin ^2a=1$

$\Rightarrow \cos a=\sin a=\pm \frac{1}{\sqrt{2}}$

b.

Vì $\sin a=\cos a=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \sin a\cos a=\frac{1}{2}$

$E=\frac{\sin a.\cos a}{\tan ^2a+\cot ^2a}=\frac{\frac{1}{2}}{1+1}=\frac{1}{4}$