Question

cot theta minus cos theta whole square equals to 1 minus cos theta upon 1 + cos theta​

Answer 1

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