Question

$prove \: that \: \: \: \: \: \: \: \\ \: 1 - \cos( \alpha) \div 1 + \cos( \alpha ) = ( \cot \alpha - \csc \alpha ) {}^{2}$
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Answer 1

Answer:

Step-by-step explanation:

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

$=cosec^2\alpha -2 \times cot\alpha \times cosec\alpha + cot^2\alpha\\= (cot\alpha-cosec\alpha)^2$