Chọn C.

Ta có: cota + tana) ² = cot²a + 2.cota.tana + tan²a

= (cot²a + 1) + (tan²a + 1)

Ta có : \(\sin^2a+\cos^2a=1\Rightarrow\cos a=\frac{\sqrt{21}}{5}\)

Ta có : \(\frac{\cot a-\tan a}{\cot a+\tan a}=\frac{\frac{\cos a}{\sin a}-\frac{\sin a}{\cos a}}{\frac{\cos a}{\sin a}+\frac{\sin a}{\cos a}}\\ =\frac{\frac{\frac{\sqrt{21}}{5}}{\frac{2}{5}}-\frac{\frac{2}{5}}{\frac{\sqrt{21}}{5}}}{\frac{\frac{\sqrt{21}}{5}}{\frac{2}{5}}+\frac{\frac{2}{5}}{\frac{\sqrt{21}}{5}}}=\frac{17}{25}=0,68\)

\(A=\dfrac{cota-tana}{tana+2\cdot cota}\)

\(=\dfrac{\dfrac{cosa}{sina}-\dfrac{sina}{cosa}}{\dfrac{sina}{cosa}+2\cdot\dfrac{cosa}{sina}}\)

\(=\dfrac{cos^2a-sin^2a}{sina\cdot cosa}:\dfrac{sin^2a+2\cdot cos^2a}{sina\cdot cosa}\)

\(=\dfrac{cos^2a-sin^2a}{sin^2a+2\cdot cos^2a}\)

\(=\dfrac{1-2\cdot sin^2a}{sin^2a+2\left(1-sin^2a\right)}\)

\(=\dfrac{1-2\cdot sin^2a}{-sin^2a+2}\)

\(=\dfrac{1-2\cdot\left(\dfrac{1}{3}\right)^2}{-\left(\dfrac{1}{3}\right)^2+2}=\dfrac{1-\dfrac{2}{9}}{-\dfrac{1}{9}+2}=\dfrac{7}{9}:\dfrac{17}{9}=\dfrac{7}{17}\)

\(-\frac{\pi}{2}< a< 0\Rightarrow cosa>0\)

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

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

\(A=\frac{tana+cota}{1+tan^2a}=\frac{tana+\frac{1}{tana}}{1+tan^2a}=\frac{1+tan^2a}{\left(1+tan^2a\right)tana}=\frac{1}{tana}=cota\)

Ta có: `sin^2 a+cos^2 a=1`

    `=>cos a=+- 4/5`   Mà `90^o < a < 180^o`

    `=>cos a=-4/5`

    `=>{(tan a=[sin a]/[cos a]=-3/4),(cot a=1/[tan a]=-4/3):}`

Có: `E=[cot a-2tan a]/[tan a+3cot a]`

      `E=[-4/3+2. 3/4]/[-3/4- 3. 4/3]=-2/57`.

\(sina=\frac{3}{5}\Rightarrow sin^2a=\frac{9}{25}\) ; \(cos^2a=1-\frac{9}{25}=\frac{16}{25}\)

\(A=\frac{cota+tana}{cota-tana}=\frac{sina.cosa\left(cota+tana\right)}{sina.cosa\left(cota-tana\right)}=\frac{cos^2a+sin^2a}{cos^2a-sin^2a}=\frac{1}{cos^2a-sin^2a}=\frac{1}{\frac{16}{25}-\frac{9}{25}}=\frac{25}{7}\)

\(B=\frac{sin^2a-cos^2a}{sin^2a-3cos^2a}=\frac{\frac{sin^2a}{sin^2a}-\frac{cos^2a}{sin^2a}}{\frac{sin^2a}{sin^2a}-\frac{3cos^2a}{sin^2a}}=\frac{1-cot^2a}{1-3cot^2a}=\frac{1-\left(-\frac{1}{3}\right)^2}{1-3\left(-\frac{1}{3}\right)^2}=\)

\(C_1=sin^2a+cos^2a+cos^2a=1+cos^2a=1+\frac{1}{1+tan^2a}=1+\frac{1}{1+\left(-2\right)^2}\)

\(C_2=\left(sin^2a+cos^2a\right)\left(sin^2a-cos^2a\right)=sin^2a-cos^2a=1-2cos^2a\)

\(=1-\frac{2}{1+tan^2a}=1-\frac{2}{1+\left(-2\right)^2}\)