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

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

\(cot^2a-cos^2a=\frac{cos^2a}{sin^2a}-cos^2a=cos^2a\left(\frac{1}{sin^2a}-1\right)=cos^2a\left(\frac{1-sin^2a}{sin^2a}\right)\)

\(=cos^2a.\frac{cos^2a}{sin^2a}=cos^2a.cot^2a\)

Câu cuối đề bài sai

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

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

\(cot^2a-cos^2a=\frac{cos^2a}{sin^2a}-cos^2a=cos^2a\left(\frac{1}{sin^2a}-1\right)=cos^2a\left(\frac{1-sin^2a}{sin^2a}\right)\)

\(=cos^2a\left(\frac{cos^2a}{sin^2a}\right)=cos^2a.cot^2a\)

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

a) \(\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos a}\)

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

\(\Leftrightarrow1-\cos^2\alpha=\sin^2\alpha\)

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

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

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

\(=\frac{cos^2a+sin^2a}{cos^2a}=\frac{1}{cos^2a}=VP\)(Đồng thời áp dụng định lý Py - ta - go)

b) \(VT=1+cot^2a=1+\frac{cos^2a}{sin^2a}\)

\(=\frac{sin^2a+cos^2a}{sin^2a}=\frac{1}{sin^2a}=VP\)(Đồng thời áp dụng định lý Py - ta - go)

vế trái =\(\frac{\sin}{1+\cot}\)+\(\frac{\cos}{1+\tan}\)= \(\frac{sin}{1+\frac{cos}{sin}}\)+\(\frac{cos}{1+\frac{sin}{cos}}\)= \(\frac{sin^2}{\sin+cos}\)+\(\frac{cos^2}{sin+cos}\)= \(\frac{sin^2+cos^2}{sin+cos}\)=\(\frac{1}{sin+cos}\)= vế phải

1. \(\frac{cos\alpha+sin\alpha}{cos\alpha-sin\alpha}=\frac{1+\frac{sin\alpha}{cos\alpha}}{1-\frac{sin\alpha}{cos\alpha}}=\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3\)

2. \(cos\beta=2sin\beta\Rightarrow cos^2\beta=4sin^2\beta\). Do \(cos^2\beta+sin^2\beta=1\Rightarrow5sin^2\beta=1\Rightarrow sin\beta=\frac{1}{\sqrt{5}}\)

\(\Rightarrow cos\beta=\frac{2}{\sqrt{5}}\). Vậy \(sin\beta.cos\beta=\frac{2}{5}\)

3. a. Nhân chéo ra được hệ thức \(sin^2\alpha+cos^2\alpha=1\)

b. Chú ý \(cot^2\alpha=\frac{cos^2\alpha}{sin^2\alpha}\)