a, ta có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)

                  \(\frac{1}{3}\)= \(\frac{\sin\alpha}{\cos\alpha}\)

                    \(\cos\alpha\)= 3 \(\sin\alpha\)

ta có \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\)= \(\frac{3\sin\alpha+\sin\alpha}{3\sin\alpha-\sin\alpha}\)= \(\frac{4\sin\alpha}{2\sin\alpha}\)= \(2\)

#mã mã#

a) Ta có A=\dfrac{\tan \alpha+3 \dfrac{1}{\tan \alpha}}{\tan \alpha+\dfrac{1}{\tan \alpha}}=\dfrac{\tan ^{2} \alpha+3}{\tan ^{2} \alpha+1}=\dfrac{\dfrac{1}{\cos ^{2} \alpha}+2}{\dfrac{1}{\cos ^{2} \alpha}}=1+2 \cos ^{2} \alphaA=tanα+tanα1​tanα+3tanα1​​=tan2α+1tan2α+3​=cos2α1​cos2α1​+2​=1+2cos2α Suy ra A=1+2 \cdot \dfrac{9}{16}=\dfrac{17}{8}A=1+2⋅169​=817​.

b) B=\dfrac{\dfrac{\sin \alpha}{\cos ^{3} \alpha}-\dfrac{\cos \alpha}{\cos ^{3} \alpha}}{\dfrac{\sin ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{3 \cos ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{2 \sin \alpha}{\cos ^{3} \alpha}}=\dfrac{\tan \alpha\left(\tan ^{2} \alpha+1\right)-\left(\tan ^{2} \alpha+1\right)}{\tan ^{3} \alpha+3+2 \tan \alpha\left(\tan ^{2} \alpha+1\right)}B=cos3αsin3α​+cos3α3cos3α​+cos3α2sinα​cos3αsinα​−cos3αcosα​​=tan3α+3+2tanα(tan2α+1)tanα(tan2α+1)−(tan2α+1)​.

Suy ra B=\dfrac{\sqrt{2}(2+1)-(2+1)}{2 \sqrt{2}+3+2 \sqrt{2}(2+1)}=\dfrac{3(\sqrt{2}-1)}{3+8 \sqrt{2}}B=22​+3+22​(2+1)2​(2+1)−(2+1)​=3+82​3(2​−1)​.

Lời giải:
a.

$\tan a+\cot a=2\Leftrightarrow \tan a+\frac{1}{\tan a}=2$

$\Leftrightarrow \frac{\tan ^2a+1}{\tan a}=2$

$\Leftrightarrow \tan ^2a-2\tan a+1=0$

$\Leftrightarrow (\tan a-1)^2=0\Rightarrow \tan a=1$

$\cot a=\frac{1}{\tan a}=1$

$1=\tan a=\frac{\cos a}{\sin a}\Rightarrow \cos a=\sin a$

Mà $\cos ^2a+\sin ^2a=1$

$\Rightarrow \cos a=\sin a=\pm \frac{1}{\sqrt{2}}$

b.

Vì $\sin a=\cos a=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \sin a\cos a=\frac{1}{2}$

$E=\frac{\sin a.\cos a}{\tan ^2a+\cot ^2a}=\frac{\frac{1}{2}}{1+1}=\frac{1}{4}$

Câu 1: 

Ta có: \(\cos\left(90^0-\alpha\right)=\sin\alpha\)

\(\Leftrightarrow\sin\alpha=1:\sqrt{\dfrac{1^2+2^2}{1}}=1:\sqrt{5}=\dfrac{\sqrt{5}}{5}\)

Câu 2: 

a) \(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5}\)

\(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)

a, ta có \(\cos^2\alpha\)+  \(\sin^2\alpha\)= 1

                  1/5 + \(\cos^2\alpha\)= 1

                               \(\cos^2\alpha\)= 4/5

\(4\cos^2\alpha\)+6 \(\sin^2\alpha\)= 4 . 4/5 + 6.1/5=22/5

b, \(\sin\alpha\)= 2/3 

\(\sin^2\alpha\)= 4/9

\(\cos^2\alpha=\frac{5}{9}\)

\(5\cos^2\alpha+2\sin^2=\frac{5.5}{9}+\frac{2.4}{9}=\frac{33}{9}\)

#mã mã#

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

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

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