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: \(\sin a+cosa=\sqrt2\)

=>\(\sqrt2\cdot\sin\left(a+\frac{\pi}{4}\right)=\sqrt2\)

=>\(\sin\left(a+\frac{\pi}{4}\right)=1\)

=>\(a+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
=>\(a=\frac{\pi}{4}+k2\pi\)

\(cosa=cos\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\)

\(\sin a=\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\)

\(\tan a=\tan\left(\frac{\pi}{4}\right)=1\)

\(\cot a=\cot\left(\frac{\pi}{4}\right)=1\)

b: \(F=\sin^5a+cos^5a\)

\(=\sin^5\left(\frac{\pi}{4}\right)+cos^5\left(\frac{\pi}{4}\right)=\left(\frac{\sqrt2}{2}\right)^5+\left(\frac{\sqrt2}{2}\right)^5\)

\(=\frac{4\sqrt2}{32}+\frac{4\sqrt2}{32}=\frac{8\sqrt2}{32}=\frac{\sqrt2}{4}\)

a: \(\sin^2a+cos^2a=1\)

=>\(cos^2a=1-\left(\frac45\right)^2=1-\frac{16}{25}=\frac{9}{25}\)

=>\(cosa=\frac35\)

\(A=5\cdot\sin a+3\cdot cosa\)

\(=5\cdot\frac45+3\cdot\frac35=4+\frac95=\frac{29}{5}\)

b:Ta có: \(\tan a\cdot\cot a=1\)

=>\(\tan a=1:\frac13=3\)

\(B=\frac{\sin a-cosa}{\sin a+cosa}\)

\(=\frac{\sin a-cosa}{cosa}:\frac{\sin a+cosa}{cosa}=\frac{\left(\tan a-1\right)}{\tan a+1}\)

\(=\frac{3-1}{3+1}=\frac24=\frac12\)

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

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

Ta có \(sin\alpha+cos\alpha=1\)=>\(cos\alpha=1-sin\alpha=1-0,6=0,4\)

\(tan\alpha=\frac{sin\alpha}{cos\alpha}=\frac{0,6}{0,4}=1,5\)

vo minh khoa \(sin^2\alpha+cos^2\alpha=1\) nhé , chỉnh lại đi

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

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

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

Do \(\pi< a< \frac{3\pi}{2}\Rightarrow cosa< 0\)

\(cosa=-\sqrt{1-sin^2a}=-\sqrt{1-0,6^2}=-\frac{4}{5}\)

\(tana=\frac{sina}{cosa}=-\frac{3}{4}\)

\(cota=\frac{1}{tana}=-\frac{4}{3}\)