a) \(\dfrac{2sina+3cosa}{3sina-4cosa}=\dfrac{9}{5}\)

b) \(\dfrac{sina.cosa}{sin^2a-sina.cosa+cos^2a}=0\)

​\(a.\dfrac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-4\cos\alpha}=\dfrac{2\left(3cos\alpha\right)+3cos\alpha}{3\left(3cos\alpha\right)-4cos\alpha}=\dfrac{9cos\alpha}{5cos\alpha}=\dfrac{9}{5}\)
\(b.\dfrac{sin\alpha cos\alpha}{sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{9cos^2\alpha-3cos^2\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{7cos^2\alpha}=\dfrac{3}{7}\)

Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}\)

Lại có: \(\dfrac{1}{cot\alpha}=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{sin^2\alpha}{cos\alpha.sin\alpha}=\dfrac{1}{\sqrt{5}}\)

\(\Rightarrow A=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}+\dfrac{sin^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

Ta có : cot α = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)

\(A=\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}\)

\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)

\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

a, Sử dụng tích chéo:

Ta có:

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

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

Mà \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Rightarrow1-\sin^2\alpha=\cos^2\alpha\)

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

Từ (1), (2)

\(\Rightarrow\)\(\cos\alpha.\cos\alpha=\)\(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)\)

\(\Rightarrow\)\(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\) (đpcm)

b/ xem lại đề

sr bạn nha mình ghi thiếu đằng sau biểu thức đó là = 4

ta có : \(A=\dfrac{sin^3\alpha+cos^3\alpha}{2sin\alpha.cos^2\alpha+cos^2\alpha.sin^2\alpha}\)

\(\Leftrightarrow A=\dfrac{\dfrac{sin^3\alpha}{cos^3\alpha}+\dfrac{cos^3\alpha}{cos^3\alpha}}{\dfrac{2sin\alpha.cos^2\alpha}{cos^3\alpha}+\dfrac{cos\alpha.sin^2\alpha}{cos^3\alpha}}=\dfrac{tan^3\alpha+1}{2tan\alpha+tan^2\alpha}\)

\(\Leftrightarrow A=\dfrac{\left(\dfrac{3}{4}\right)^3+1}{2\left(\dfrac{3}{4}\right)+\left(\dfrac{3}{4}\right)^2}=\dfrac{91}{132}\)

a) Cần chứng minh \(\dfrac{1-cos\alpha}{sin\alpha}=\dfrac{sin\alpha}{1+cos\alpha}\)

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

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

Giả sử tam giác ABC vuông tại A

Ta có: \(\left\{{}\begin{matrix}sin^2B=\dfrac{AC^2}{BC^2}\\cos^2B=\dfrac{AB^2}{BC^2}\end{matrix}\right.\Rightarrow sin^2B+cos^2B=\dfrac{AC^2+AB^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\)

a)\(\dfrac{1-cosa}{sina}=\dfrac{sina}{1+cosa}\)

<=>\(\left(1-cosa\right)\left(1+cosa\right)=sin^2a\)

<=>\(1-cos^2a=sin^2a\) (lđ)

b)Ta có VT=\(\dfrac{cosa}{1+sina}+tga=\dfrac{cosa}{1+sina}+\dfrac{sina}{cosa}=\dfrac{cos^2a+sin^2a+sina}{\left(1+sina\right)cosa}=\dfrac{1+sina}{\left(1+sina\right)cosa}=\dfrac{1}{cosa}=vp\left(dpcm\right)\)

Ta có:

\(sin=\dfrac{doi}{huyen}\); \(cos=\dfrac{ke}{chuyen}\);\(tan=\dfrac{doi}{ke}\); \(cot=\dfrac{ke}{doi}\)

Dùng cái này làm được hết mấy câu đó.

nếu bn thấy dùng cách của hùng có hới dài thì bn chỉ cần sử dụng cách đó cho 3 ý trên thôi . còn 3 ý dưới bn có thể sử dụng công thức \(sin^2x+cos^2x=1\) vừa chứng minh xong để giải quyết .

\(\dfrac{sina+cosa}{sina-cosa}=\dfrac{\dfrac{sina+cosa}{cosa}}{\dfrac{sina-cosa}{cosa}}=\dfrac{tana+1}{tana-1}=\dfrac{3}{1}=3\)

Có \(\dfrac{sin\alpha}{cos\alpha}=tan\alpha=2\)\(\Rightarrow sin\alpha=2cos\alpha\)

\(\dfrac{sin\alpha+cos\alpha}{sin\alpha-cos\alpha}=\dfrac{2cos\alpha+cos\alpha}{2cos\alpha-cos\alpha}=\dfrac{3cos\alpha}{cos\alpha}=3\)

\(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\\ VT=\dfrac{sin^2a+2sinacosa+cos^2a-sin^2a+2sinacosa-cos^2a}{sinacosa}\\ =\dfrac{4sinacosa}{sinacosa}=4=VP\)

a: \(S=cos^2a\left(1+tan^2a\right)=cos^2a\cdot\dfrac{1}{cos^2a}=1\)

b: \(VP=\dfrac{1+sin2a-1+sin2a}{\dfrac{1}{2}\cdot sin2a}=\dfrac{2\cdot sin2a}{\dfrac{1}{2}\cdot sin2a}=4=VT\)

sữa đề chút nha :

+) ta có : \(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{\left(sin\alpha+cos\alpha\right)^2}{\left(sin\alpha+cos\alpha\right)\left(cos\alpha-sin\alpha\right)}=\dfrac{sin\alpha+cos\alpha}{cos\alpha-sin\alpha}\)

+) ta có :

\(B=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)

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

\(=1-3sin^2\alpha.cos^2\alpha+3sin^2\alpha.cos^2\alpha=1\)