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

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

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

a.\(1-\sin^2\alpha=\cos^2\alpha\)

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

c.\(\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=1-\cos^2\alpha=\sin^2\alpha\)

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

e.\(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha=\tan^2\alpha\left(1-\sin^2\alpha\right)=\tan^2\alpha.\cos^2\alpha=\sin^2\alpha\)

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

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

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

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

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

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

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

hỏi tí chớ \(TanB=2\) hay \(Tan\alpha=2\) vậy .

a) ta có : \(VT=sin^2\alpha+cos^2\alpha=\left(\dfrac{đối}{huyền}\right)^2+\left(\dfrac{kề}{huyền}\right)^2\)

\(=\dfrac{\left(đối\right)^2+\left(kề\right)^2}{\left(huyền\right)^2}=\dfrac{\left(huyền\right)^2}{\left(huyền\right)^2}=1=VP\left(đpcm\right)\)

b) ta có : \(VP=cot^2\alpha-cos^2\alpha=\dfrac{cos^2\alpha}{sin^2\alpha}-cos^2\alpha=cos^2\alpha\left(\dfrac{1}{sin^2\alpha}-1\right)\)

\(=cos^2\alpha\left(\dfrac{1-sin^2\alpha}{sin^2\alpha}\right)=cos^2\alpha\dfrac{cos^2\alpha}{sin^2\alpha}=cos^2\alpha.cot^2\alpha=VT\left(đpcm\right)\)

Những biểu thức này đều không tính toán ra được giá trị cụ thể nên không phù hợp với yêu cầu "tính". Mình nghĩ bạn nên xem xét lại yêu cầu đề.

Lời giải:

Biểu thức $A$ dạng như vậy là gọn rồi bạn ạ. Biến đổi thêm cũng không có ý nghĩa.

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\(B=\sin ^2a+\sin 2a-3\cos ^3a\)

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

\(=\frac{\tan a-1-\cot a}{2}\)

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

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

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

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

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