Ta có: \(tan\alpha=2\Leftrightarrow\dfrac{sin\alpha}{cos\alpha}=2\Leftrightarrow sin\alpha=2cos\alpha\)

A = \(\dfrac{16cos^2\alpha+6cos^2\alpha}{20cos^2\alpha-2cos^2\alpha}=\dfrac{22cos^2\alpha}{18cos^2\alpha}=\dfrac{11}{9}\)

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

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

\(P=\dfrac{2sin\alpha-3cos\alpha}{3sin\alpha+2cos\alpha}\\ =\dfrac{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}\\ =\dfrac{2tan\alpha-3}{3tan\alpha+2}=\dfrac{2.3-3}{3.3+2}=\dfrac{3}{11}\)

Ta có: \(1 + {\tan ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }}\quad (\alpha  \ne {90^o})\)

\( \Rightarrow \frac{1}{{{{\cos }^2}\alpha }} = 1 + {3^2} = 10\)

\( \Leftrightarrow {\cos ^2}\alpha  = \frac{1}{{10}} \Leftrightarrow \cos \alpha  =  \pm \frac{{\sqrt {10} }}{{10}}\)

Vì \({0^o} < \alpha  < {180^o}\) nên \(\sin \alpha  > 0\).

Mà \(\tan \alpha  = 3 > 0 \Rightarrow \cos \alpha  > 0 \Rightarrow \cos \alpha  = \frac{{\sqrt {10} }}{{10}}\)

Lại có: \(\sin \alpha  = \cos \alpha .\tan \alpha  = \frac{{\sqrt {10} }}{{10}}.3 = \frac{{3\sqrt {10} }}{{10}}.\)

\( \Rightarrow P = \dfrac{{2.\frac{{3\sqrt {10} }}{{10}} - 3.\frac{{\sqrt {10} }}{{10}}}}{{3.\frac{{3\sqrt {10} }}{{10}} + 2.\frac{{\sqrt {10} }}{{10}}}} = \dfrac{{\frac{{\sqrt {10} }}{{10}}\left( {2.3 - 3} \right)}}{{\frac{{\sqrt {10} }}{{10}}\left( {3.3 + 2} \right)}} = \dfrac{3}{{11}}.\)

\(A=\dfrac{cos^2a-sin^2a}{\dfrac{cos^2a}{sin^2a}-\dfrac{sin^2a}{cos^2a}}-cos^2a=\dfrac{cos^2a.sin^2a\left(cos^2a-sin^2a\right)}{\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)}-cos^2a\)

\(=cos^2a.sin^2a-cos^2a=cos^2a\left(sin^2a-1\right)=-cos^4a\)

\(B=\sqrt{\left(1-cos^2a\right)^2+6cos^2a+3cos^4a}+\sqrt{\left(1-sin^2a\right)^2+6sin^2a+3sin^4a}\)

\(=\sqrt{4cos^4a+4cos^2a+1}+\sqrt{4sin^4a+4sin^2a+1}\)

\(=\sqrt{\left(2cos^2a+1\right)^2}+\sqrt{\left(2sin^2a+1\right)^2}\)

\(=2\left(sin^2a+cos^2a\right)+2=4\)

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 

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

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

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

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

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

OK

\(A=\frac{\frac{sina}{cos^3a}-\frac{cosa}{cos^3a}}{tan^3a+3+\frac{2sina}{cos^3a}}=\frac{tana.\frac{1}{cos^2a}-\frac{1}{cos^2a}}{tan^3a+3+2tana.\frac{1}{cos^2a}}\)

\(=\frac{tana\left(1+tan^2a\right)-\left(1+tan^2a\right)}{tan^3a+3+2tana\left(1+tan^2a\right)}=\frac{3\left(1+9\right)-\left(1+9\right)}{27+3+2.3.\left(1+9\right)}=...\)

Cách 1:

Ta có: \(tan\alpha=\sqrt{2}\Rightarrow\left\{{}\begin{matrix}\dfrac{sin\alpha}{cos\alpha}=\sqrt{2}\\1+\left(\sqrt{2}\right)^2=\dfrac{1}{cos^2\alpha}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=\sqrt{2}\cdot cos\alpha\\cos^2\alpha=\dfrac{1}{3}\end{matrix}\right.\)

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

    \(=\dfrac{\sqrt{2}\cdot cos\alpha-cos\alpha}{\left(\sqrt{2}\cdot cos\alpha\right)^3+3cos^3\alpha+2\cdot\sqrt{2}\cdot cos\alpha}\)

    \(=\dfrac{cos\alpha\left(\sqrt{2}-1\right)}{2\sqrt{2}\cdot cos^3\alpha+3cos^3\alpha+2\sqrt{2}\cdot cos\alpha}\)

    \(=\dfrac{cos\alpha\left(\sqrt{2}-1\right)}{cos\alpha\left(2\sqrt{2}\cdot cos^2\alpha+3cos^2\alpha+2\sqrt{2}\right)}\)

    \(=\dfrac{\sqrt{2}-1}{2\sqrt{2}\cdot cos^2\alpha+3cos^2\alpha+2\sqrt{2}}\)

Thay \(cos^2\alpha=\dfrac{1}{3}\) vào \(P\) ta có:

\(P=\dfrac{\sqrt{2}-1}{2\sqrt{2}\cdot\dfrac{1}{3}+3\cdot\dfrac{1}{3}+2\sqrt{2}}=\dfrac{\sqrt{2}-1}{1+\dfrac{8}{3}\sqrt{2}}\)

    \(=\dfrac{3\left(\sqrt{2}-1\right)}{3\left(1+\dfrac{8}{3}\sqrt{2}\right)}=\dfrac{3\left(\sqrt{2}-1\right)}{3+8\sqrt{2}}\)

    \(=\dfrac{3\left(\sqrt{2}-1\right)}{3+2^3\sqrt{2}}=\dfrac{a\left(\sqrt{b}-1\right)}{a+b^3\sqrt{b}}\)

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\Rightarrow a+b=5\)

Chọn đáp án A.

Cách 2:

\(P=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}=\dfrac{\left(sin\alpha-cos\alpha\right)\div cos^3\alpha}{\left(sin^3\alpha+3cos^3\alpha+2sin\alpha\right)\div cos^3\alpha}\)

    \(=\dfrac{\dfrac{sin\alpha}{cos^3\alpha}-\dfrac{1}{cos^2\alpha}}{\dfrac{sin^3\alpha}{cos^3\alpha}+3+2\cdot\dfrac{sin\alpha}{cos^3\alpha}}=\dfrac{\dfrac{sin\alpha}{cos\alpha}\cdot\dfrac{1}{cos^2\alpha}-\dfrac{1}{cos^2\alpha}}{tan^3\alpha+3+2\cdot\dfrac{sin\alpha}{cos\alpha}\cdot\dfrac{1}{cos^2\alpha}}\)

    \(=\dfrac{tan\alpha\cdot\left(1+tan^2\alpha\right)-\left(1+tan^2\alpha\right)}{tan^3\alpha+3+2tan\alpha\cdot\left(1+tan^2\alpha\right)}\)

Thay \(tan\alpha=\sqrt{2}\) vào ta có:

\(P=\dfrac{\sqrt{2}\cdot\left[1+\left(\sqrt{2}\right)^2\right]-\left[1+\left(\sqrt{2}\right)^2\right]}{\left(\sqrt{2}\right)^3+3+2\sqrt{2}\cdot\left[1+\left(\sqrt{2}\right)^2\right]}=\dfrac{3\sqrt{2}-3}{2\sqrt{2}+3+6\sqrt{2}}\)

    \(=\dfrac{3\left(\sqrt{2}-1\right)}{3+8\sqrt{2}}=\dfrac{3\left(\sqrt{2}-1\right)}{3+2^3\sqrt{2}}=\dfrac{a\left(\sqrt{b}-1\right)}{a+b^3\sqrt{b}}\)

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\Rightarrow a+b=3+2=5\)

Chọn đáp án A

\(A=\dfrac{3sin\alpha-cos\alpha}{sin\alpha+cos\alpha}=\dfrac{\dfrac{3sin\alpha}{cos\alpha}-1}{\dfrac{sin\alpha}{cos\alpha}-1}=\dfrac{3tan\alpha-1}{tan\alpha-1}\)\(=\dfrac{3\sqrt{2}-1}{\sqrt{2}-1}=5+2\sqrt{2}\).