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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 )
\(\Rightarrow\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\)
\(\frac{sin^2\alpha}{cos\alpha.\left(1+\frac{sin\alpha}{cos\alpha}\right)}-\frac{cos^2\alpha}{sin\alpha.\left(1+\frac{cos\alpha}{sin\alpha}\right)}=\frac{sin^2\alpha}{cos\alpha+sin\alpha}-\frac{cos^2\alpha}{sin\alpha+cos\alpha}=\frac{\left(sin\alpha+cos\alpha\right).\left(sin\alpha-cos\alpha\right)}{sin\alpha+cos\alpha}=sin\alpha-cos\alpha\)
\(a,1-sin^2\alpha=cos^2\alpha\)
\(b,\left(1-cos\alpha\right)\left(1+cos\alpha\right)=1-cos^2\alpha=sin^2\alpha\)
\(c,1+sin^2\alpha+cos^2\alpha=1+1=2\)
\(d,sin\alpha-sin\alpha.cos^2\alpha=sin\alpha.\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)
\(e,sin^2\alpha+cos^2\alpha+2sin^2\alpha.cos^2\alpha\)
\(=1+2sin^2\alpha.cos^2\alpha\)
a)\(tan3A=tan\left(A+2A\right)\)
\(=\frac{tanA+tan2A}{1-tanAtan2A}\)
\(=\frac{\frac{tanA+2tanA}{1-tan^2A}}{\frac{1-2tan^2A}{1-tan^2A}}\)
\(=\frac{\left(tanA-tan^3A+2tanA\right)}{1-tan^2A-2tan^2A}\)
\(=\frac{3tanA-tan^3A}{1-3tan^2A}\)
b)\(VT=cos^6A+sin^6A\)
\(=\left(cos^2A\right)^3+\left(sin^2A\right)^3\)
\(=\left(cos^2A+sin^2A\right)^3-3cos^2Asin^2A\left(cos^2A+sin^2A\right)^2\)
\(=1^3-3cos^2Asin^2A\left(1\right)^2\).Từ đó,\(sin^2A+cos^2A=1\)
\(=1-3cos^2Asin^2A=VP\)
phần b tui sai