A

Chọn A

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

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

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! 

tan a=2 nên sina/cosa=2

=>sina=2cosa

\(A=\dfrac{sinacosa\left(sin^2a+cos^2a\right)}{\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a}=\dfrac{sina\cdot cosa}{1-2\cdot\left(sina\cdot cosa\right)^2}\)

\(=\dfrac{2cosa\cdot cosa}{1-2\cdot\left(2cosa\cdot cosa\right)^2}=\dfrac{2cos^2a}{1-8cos^2a}\)

a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)

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

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

b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)

\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)

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

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

c/ \(C=sin^6x+cos^6x+3sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3sin^2x.cos^2x\)

\(=sin^4x-sin^2x.cos^2x+cos^4x+3sin^2x.cos^2x\)

\(=sin^4x+cos^4x+2sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2=1\)

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

mình làm r nha

https://hoc24.vn/cau-hoi/biet-cotadfrac12-gia-tri-bieu-thuc-adfrac4sinalpha5cosalpha2sinalpha-3cosalpha-bang-bao-nhieughi-ro-tung-loi-giai-nha.5724337531039

a) \(\cos^4\alpha-\sin^4\alpha=\left(\cos^2\alpha+\sin^2\alpha\right)\left(\cos^2\alpha-\sin^2\alpha\right)=\cos^2\alpha-\sin^2\alpha\)

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

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

\(\Leftrightarrow\)\(1-\left(\sin^2\alpha+\cos^2\alpha\right)=0\)\(\Leftrightarrow\)\(1-1=0\) ( luôn đúng ) 

c) \(\frac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha.\cos\alpha}=\frac{2\cos\alpha.2\sin\alpha}{\sin\alpha.\cos\alpha}=4\)

um, hình như câu b) chỗ 1-.... đó hơi sai nếu viết từ bước trên xuống á bạn!

mình nghĩ là: sau dấu bằng đầu tiên, sau đó là:

\(=cos^2\alpha=1-sin^2\alpha\)(luôn đúng)

CẢM ƠN bạn nhiều lắm luôn nha!!!!!