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)

Ta có:

\({\cos ^4}\alpha {\sin ^4}\alpha  = \left( {{{\cos }^2}\alpha  - {{\sin }^2}\alpha } \right)\left( {{{\cos }^2}\alpha  + {{\sin }^2}\alpha } \right) \\= {\cos ^2}\alpha  - {\sin ^2}\alpha = {\cos ^2}\alpha  - (1 - {\cos ^2}\alpha ) \\= {\cos ^2}\alpha  - 1 + {\cos ^2}\alpha  = 2{\cos ^2}\alpha  - 1\)

(đpcm)

b)

Ta có:

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

(đpcm)

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

a)    Ta có:

\(\begin{array}{l}{\sin ^4}\alpha  - {\cos ^4}\alpha  = 1 - 2{\cos ^2}\alpha \\ \Leftrightarrow \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)\left( {{{\sin }^2}\alpha  - {{\cos }^2}\alpha } \right) = 1 - 2{\cos ^2}\alpha \\ \Leftrightarrow {\sin ^2}\alpha  - {\cos ^2}\alpha  - 1 + 2{\cos ^2}\alpha  = 0\\ \Leftrightarrow {\sin ^2}\alpha  + {\cos ^2}\alpha  - 1 = 0\\ \Leftrightarrow 1 - 1 = 0\\ \Leftrightarrow 0 = 0\end{array}\)

Đẳng thức luôn đúng

b)    Ta có:

\(\begin{array}{l}\tan \alpha  + \cot \alpha  = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{\cos \alpha .\sin \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{1}{{\sin \alpha .\cos \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\end{array}\)

Đẳng thức luôn đúng

1) \(1-2\sin\alpha.\cos\alpha=\sin^2\alpha-2\sin\alpha.\cos\alpha+\cos^2\alpha=\left(\sin\alpha-\sin\alpha\right)^2\ge0\)

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

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