\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)

\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)

\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)

\(0< a< 90\Rightarrow sina>0\Rightarrow sina=\sqrt{1-cos^2a}=\frac{3}{5}\)

\(A=\frac{cosa+sina}{cosa-sina}=\frac{\frac{4}{5}+\frac{3}{5}}{\frac{4}{5}-\frac{3}{5}}=7\)

`A=sin(π-α)+cos(π+α)+cos(-α)`

`= sinα-cosα+cosα=sinα=3/5`

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

b) \(\sin x+\cos x=\dfrac{3}{2}\)

\(\left(\sin x+\cos x\right)^2=\dfrac{1}{4}\)

\(\sin^2x+\cos^2x+2\sin x\cos x=\dfrac{1}{4}\)

\(2\sin x\cos x=-\dfrac{3}{4}=\sin2x\)

ý a,

ý c

Ta có:

\({\cos ^2}a + {\sin ^2}a = 1 \Rightarrow \sin a =  \pm \frac{4}{5}\)

Do \(0 < a < \frac{\pi }{2} \Leftrightarrow \sin a = \frac{4}{5}\)

\(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{4}{3}\)

Ta có;

\(\begin{array}{l}\sin \left( {a + \frac{\pi }{6}} \right) = \sin a.\cos \frac{\pi }{6} + \cos a.\sin \frac{\pi }{6} = \frac{4}{5}.\frac{{\sqrt 3 }}{2} + \frac{3}{5}.\frac{1}{2} = \frac{{3 + 4\sqrt 3 }}{{10}}\\\cos \left( {a - \frac{\pi }{3}} \right) = \cos a.\cos \frac{\pi }{3} + \sin a.\sin \frac{\pi }{3} = \frac{3}{5}.\frac{1}{2} + \frac{4}{5}.\frac{{\sqrt 3 }}{2} = \frac{{3 + 4\sqrt 3 }}{{10}}\\\tan \left( {a + \frac{\pi }{4}} \right) = \frac{{\tan a + \tan \frac{\pi }{4}}}{{1 - \tan a.tan\frac{\pi }{4}}} = \frac{{\frac{4}{3} + 1}}{{1 - \frac{4}{3}}} =  - 7\end{array}\)

1. Ta có \(\tan a=3\Rightarrow\frac{\sin a}{\cos a}=3\Rightarrow\sin a=3\cos a\)

Vậy \(\frac{\cos a+\sin a}{\cos a-\sin a}=\frac{\cos a+3\cos a}{\cos a-3\cos a}=\frac{4\cos a}{-2\cos a}=-2\)

2.Ta có \(\sin^2a+\cos^2a=1\Rightarrow\cos^2a=1-\sin^2a=1-\frac{4}{9}=\frac{5}{9}\)

\(\Rightarrow\orbr{\begin{cases}\cos a=\frac{\sqrt{5}}{3}\\\cos a=\frac{-\sqrt{5}}{3}\end{cases}}\)

Với \(\cos a=\frac{\sqrt{5}}{3}\Rightarrow\tan a=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}=\frac{2\sqrt{5}}{5}\Rightarrow\cot a=\frac{1}{\tan a}=\frac{\sqrt{5}}{2}\)

Với \(\cos a=\frac{-\sqrt{5}}{2}\Rightarrow\tan a=\frac{-2\sqrt{5}}{5}\Rightarrow\cot a=-\frac{\sqrt{5}}{2}\)

3. 

Theo hệ thức  lượng trong tam giác vuông ta có \(AB^2=BH.BC\Leftrightarrow10^2=5.BC\Rightarrow BC=20\left(cm\right)\)

Theo định lí Pitago thì \(AC=\sqrt{BC^2-AB^2}=\sqrt{20^2-10^2}=10\sqrt{3}\left(cm\right)\)

Ta có \(\tan B=\frac{AC}{AB}=\frac{10\sqrt{3}}{10}=\sqrt{3};\tan C=\frac{AB}{AC}=\frac{1}{\sqrt{3}}\)

Vậy \(\tan B=3\tan C\)