\(\dfrac{sina}{sina-cosa}-\dfrac{cosa}{cosa-sina}=\dfrac{sina+cosa}{sina-cosa}=\dfrac{1+cota}{1-cota}=\dfrac{\left(1+cota\right)^2}{1-cot^2a}\)

Đề bài ko đúng

a) Có: `1+tan^2a=1/(cos^2a)`

`<=> 1+(3/5)^2=1/(cos^2a)`

`=> cosa=\sqrt10/4`

`=> sina = \sqrt(1-cos^2a) = \sqrt6/4`

b) Có: `sin^2a + cos^2a=1`

`<=> sin^2a + (1/4)^2=1`

`=> sina=\sqrt15/4`

`=> tana = (sina)/(cosa) = \sqrt15`

a) Giả sử tam giác ABC vuông tại B có \(tanA=\dfrac{3}{5}\)

\(\Rightarrow\dfrac{BC}{AB}=\dfrac{3}{5}\Rightarrow BC=\dfrac{3}{5}AB\Rightarrow AC=\sqrt{AB^2+\dfrac{9}{25}AB^2}=\dfrac{\sqrt{34}}{5}AB\)

\(\Rightarrow\dfrac{AB}{AC}=\dfrac{5}{\sqrt{34}}\Rightarrow cosA=\dfrac{5}{\sqrt{34}}\)

\(AC=\dfrac{\sqrt{34}}{5}AB\Rightarrow AC=\dfrac{\sqrt{34}}{5}.\dfrac{5}{3}BC=\dfrac{\sqrt{34}}{3}BC\Rightarrow\dfrac{BC}{AC}=\dfrac{3}{\sqrt{34}}\)

\(\Rightarrow sinA=\dfrac{3}{\sqrt{34}}\)

b) cũng tương tự như câu a thôi,bạn tự tính nha

b) khai triển hằng đẳng thức là ra

a) nhân tích chéo

\(\frac{\cos\alpha}{1-\sin\alpha}=\frac{1+\sin\alpha}{\cos\alpha}\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)\(\Leftrightarrow\cos^2\alpha+\sin^2\alpha=1\)(luôn đúng)

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

\(=\frac{4\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}=4\)(đpcm)

4

a: \(\sin^2a+\cos^2a=1\)

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

hay \(\dfrac{\cos a}{1-\sin a}=\dfrac{1+\sin a}{\cos a}\)

b: \(VT=\dfrac{\left(\sin a+\cos a+\sin a-\cos a\right)\left(\sin a+\cos a-\sin a+\cos a\right)}{\sin a\cdot\cos a}\)

\(=\dfrac{2\cdot\cos a\cdot2\sin a}{\sin a\cdot\cos a}=4\)

VT=\(\dfrac{c\text{os}a}{1-sina}\)

\(=\dfrac{c\text{os}a\left(1+sina\right)}{\left(1-sina\right)\left(1+sina\right)}=\dfrac{c\text{os}a\left(1+sina\right)}{1-sin^2a}\\ \\ \\ =\dfrac{c\text{os}a\left(1+sina\right)}{c\text{os}^2a}=\dfrac{1+sina}{c\text{os}a}=VP\left(\text{đ}pcm\right)\)

Giả sử các biểu thức đều xác định

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

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

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

Chứng minh các hằng đẳng thức trên