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

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

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

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

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

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

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

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

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

Lời giải:

Sử dụng các công thức sau:

\(\bullet \tan \alpha=\frac{1}{\cot \alpha}\)

\(\bullet \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan\alpha.\tan \beta}\)

Ta có:

\(\text{VT}=\frac{1}{\tan a+\tan b}-\frac{1}{\cot a+\cot b}=\frac{1}{\tan a+\tan b}-\frac{1}{\frac{1}{\tan a}+\frac{1}{\tan b}}\)

\(=\frac{1}{\tan a+\tan b}-\frac{\tan a\tan b}{\tan a+\tan b}=\frac{1-\tan a\tan b}{\tan a+\tan b}\)

\(=\frac{1}{\frac{\tan a+\tan b}{1-\tan a\tan b}}=\frac{1}{\tan (a+b)}=\cot (a+b)=\text{VP}\)

Ta có đpcm.

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`.

\(\sin^2\widehat{A}+\cos^2\widehat{A}=1\Leftrightarrow\cos^2\widehat{A}=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\\ \Leftrightarrow\cos\widehat{A}=\dfrac{4}{5}\\ \tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}=\dfrac{3}{4}\\ \Rightarrow\cot\widehat{A}=\dfrac{1}{\tan\widehat{A}}=\dfrac{4}{3}\)

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

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

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

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

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

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