Question

C/m các hệ thứ sau

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

b, \(\tan^2\alpha-\sin^2\alpha=\tan^2\alpha\times\sin^2\alpha\)

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Answer 1

a) ta có : \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow sin^2\alpha=1-cos^2\alpha\)

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

b) ta có : \(tan^2\alpha-sin^2\alpha=sin^2\alpha\left(\dfrac{1}{cos^2\alpha}-1\right)=sin^2\alpha\left(\dfrac{1-cos^2\alpha}{cos^2\alpha}\right)\)

\(=sin^2\alpha.\dfrac{sin^2\alpha}{cos^2\alpha}=sin^2\alpha.tan^2\alpha\left(đpcm\right)\)