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

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

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

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

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 )

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

1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)

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

\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)

\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)

1: \(cota=\sqrt{5}\)

=>\(cosa=\sqrt{5}\cdot sina\)

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

=>\(\dfrac{1}{sin^2a}=1+5=6\)

=>\(sin^2a=\dfrac{1}{6}\)

\(C=sin^2a-sina\cdot\sqrt{5}\cdot sina+\left(\sqrt{5}\cdot sina\right)^2\)

\(=sin^2a\left(1-\sqrt{5}+5\right)=\dfrac{1}{6}\cdot\left(6-\sqrt{5}\right)\)

2: tan a=3

=>sin a=3*cosa 

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

=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)

\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)

\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)

\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)

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)

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.

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

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

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

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

\(\Leftrightarrow1-2sin\alpha cos\alpha=1-2sin\alpha cos\alpha\left(đpcm\right)\)

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

2) \(cos^4\alpha+sin^2\alpha\cdot cos^2\alpha+sin^2\alpha\)

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

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

\(=cos^2\alpha+sin^2\alpha=1\)(đpcm)