\(tan\alpha=\dfrac{1}{3}\Rightarrow\dfrac{sin\alpha}{cos\alpha}=\dfrac{1}{3}\Rightarrow cos\alpha=3sin\alpha\)

Thay cosa=3sina vào A, được:

\(A=\dfrac{sin^2a+9sin^2a}{sin^2a+9sin^2a+6sin^2a}=\dfrac{10sin^2a}{16sin^2a}=\dfrac{5}{8}\)

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

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/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)

\(=\frac{\frac{cos^2a}{sin^2a}-cos^2a}{\frac{cos^2a}{sin^2a}}-\frac{sina.cosa}{\frac{cosa}{sina}}\)

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

b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)

\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)

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

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

c/ \(C=sin^6x+cos^6x+3sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3sin^2x.cos^2x\)

\(=sin^4x-sin^2x.cos^2x+cos^4x+3sin^2x.cos^2x\)

\(=sin^4x+cos^4x+2sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2=1\)

Ta có:

\(sin^2a+cos^2a=1\Leftrightarrow sin^2a+\left(\frac{1}{3}\right)^2=1\Leftrightarrow sin^2a=\frac{8}{9}\Rightarrow sina=\frac{2\sqrt{2}}{3}.\)

\(B=\frac{sin\alpha-3cosa}{sina+2cosa}=\frac{\frac{2\sqrt{2}}{3}-3.\frac{1}{3}}{\frac{2\sqrt{2}}{3}+2.\frac{1}{3}}=\frac{7-5\sqrt{2}}{2}\)

A

Chọn A

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

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

\(\frac{\cos a-\sin a}{cosa+sina}=\frac{\frac{cosa}{cosa}-\frac{sina}{cosa}}{\frac{cosa}{cosa}+\frac{sina}{cosa}}\)(chia ca tu va mau cho cosa)

                \(=\frac{1-tana}{1+tana}=vt\left(dpcm\right)\)