\(A=\dfrac{\dfrac{3sina}{sina}-\dfrac{cosa}{sina}}{\dfrac{2sina}{sina}+\dfrac{cosa}{sina}}=\dfrac{3-cota}{2+cota}=\dfrac{3-3}{2+3}=0\)

\(B=\dfrac{\dfrac{sin^2a}{sin^2a}-\dfrac{3sina.cosa}{sin^2a}+\dfrac{2}{sin^2a}}{\dfrac{2sin^2a}{sin^2a}+\dfrac{sina.cosa}{sin^2a}+\dfrac{cos^2a}{sin^2a}}=\dfrac{1-3cota+2\left(1+cot^2a\right)}{2+cota+cot^2a}=\dfrac{1-3.3+2\left(1+3^2\right)}{2+3+3^2}=...\)

a. \(A=\dfrac{3sin\alpha-cos\alpha}{2sin\alpha+cos\alpha}=\dfrac{3\dfrac{sin\alpha}{cos\alpha}-1}{2\dfrac{sin\alpha}{cos\alpha}+1}=\dfrac{3.\dfrac{1}{3}-1}{2.\dfrac{1}{3}+1}=0\)

b.\(B=\dfrac{sin^2\alpha-3sin\alpha.cos\alpha+2}{2sin^2\alpha+sin\alpha.cos\alpha+cos^2\alpha}\)\(=\dfrac{1-\dfrac{3cos\alpha}{sin\alpha}+\dfrac{2}{sin^2\alpha}}{2+\dfrac{cos\alpha}{sin\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}}=\dfrac{1-3.3+\dfrac{2}{sin^2\alpha}}{2+3+3^2}\)

Mà \(\dfrac{cos\alpha}{sin\alpha}=3,cos^2\alpha+sin^2\alpha=1\Rightarrow sin^2\alpha=\dfrac{1}{10}\)

\(B=\dfrac{1-3.3+\dfrac{2}{\dfrac{1}{10}}}{2+3+3^2}=\dfrac{6}{7}\)

Dạ em cảm ơn thầy và mọi người ạ! 

a: \(VT=\dfrac{\left(sina+cosa\right)^3-3\cdot sina\cdot cosa\left(sina+cosa\right)}{sina+cosa}\)

=(sina+cosa)^2-3*sina*cosa

=sin^2a+cos^2a-sina*cosa

=1-sina*cosa=VP

c: VT=(sin^2a+cos^2a)^2-2*sin^2a*cos^2a-(sin^2a+cos^2a)^3+3*sin^2a*cos^2a*(sin^2a+cos^2a)

=1-2sin^2a*cos^2a-1+3*sin^2a*cos^2a

=sin^2a*cos^2a=VP

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

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

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

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

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

\(1+\sin^2\alpha+\cos^2\alpha=1+1=2\)

\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha\cdot\cos^2\alpha\\ =\left(\sin^2\alpha\right)^2+2\sin^2\alpha\cdot\cos^2\alpha+\left(\cos^2\alpha\right)^2\\ =\left(\sin^2\alpha+\cos^2\alpha\right)^2\\ =1^2=1\)

\(\tan^2\alpha-\sin^2\alpha\cdot\tan^2\alpha\\ =\tan^2\alpha\left(1-\sin^2\alpha\right)\\ =\left(\frac{\sin\alpha}{\cos\alpha}\right)^2\cdot\cos^2\alpha\\ =\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\cos^2\alpha\\ =\sin^2\alpha\)

\(\cos^2\alpha+\tan^2\alpha\cdot\cos^2\alpha\\ =\cos^2\alpha+\left(\frac{\sin\alpha}{\cos\alpha}\right)^2\cdot\cos^2\alpha\\ =\cos^2\alpha+\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\cos^2\alpha\\ =\cos^2\alpha+\sin^2\alpha\\ =1\)

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

Đề bài ko chính xác, biểu thức này không rút gọn được (có thể coi việc biến đổi khả dĩ duy nhất \(1+2sina.cosa=\left(sina+cosa\right)^2\) không phải là hành động rút gọn)

chỉnh lại đề 1 chút: \(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{cos^2\alpha+sin^2\alpha+2sin\alpha.cos\alpha}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}\)

\(=\dfrac{\left(cos\alpha+sin\alpha\right)^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{cos\alpha+sin\alpha}{cos\alpha-sin\alpha}\)

Lời giải:
\(A=(\sin ^2a)^3+(\cos ^2a)^3+3\sin ^2a\cos ^2a(\sin ^2a+\cos ^2a)\)

\(=(\sin ^2a+\cos ^2a)^3=1^3=1\)

\(B=(\cos ^2a+\sin ^2a-2\sin a\cos a)+(\cos ^2a+\sin ^2a+2\sin a\cos a)\)

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

\(C=\frac{(\cos ^2a+\sin ^2a-2\sin a\cos a)-(\cos ^2a+\sin ^2a+2\sin a\cos a)}{\sin a\cos a}=\frac{(1-2\sin a\cos a)-(1+2\sin a\cos a)}{\sin a\cos a}\)

$=\frac{-4\sin a\cos a}{\sin a\cos a}=-4$

\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)

\(=tan^2a+1=\frac{1}{cos^2a}\)

\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)

\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)

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

Chọn B