Question

. Solve this please help me​

Answer 1

$\huge \red{ \underline{ question}} \\ \\ \implies \: prove \: that \: \: \: \: \frac{1 - cos \theta}{1 + cos \theta} = {(cosec \theta - cot \theta)}^{2} \\ \\ \\ \green{ \underline{proof}} \implies \\ \\ \underline{step \: by \: step \: explanation} \\ \\ \green{firstly \: taking \: left \: hand \: side} \\ \\ \implies \: \frac{1 - cos \theta}{1 + cos \theta} \\ \\ rationalising \: by \: (1 - cos \theta) \: \\ \blue{ it \: become }\\ \\ \implies \: \bigg( \frac{1 - cos \theta}{1 + cos \theta} \bigg) \times \bigg( \frac{1 - cos \theta}{1 - cos \theta} \bigg) \\ \\ using \: identities \\ \red{ \star } \implies\: {a}^{2} - {b}^{2} = (a - b)(a + b) \\ \red{ \star} \implies 1 - {cos}^{2} \theta = {sin}^{2} \theta \\ \red{\star}\implies 1/sin\theta=cosec\theta \\ \red{\star} \implies cos\theta/sin\theta=cot\theta \\ \\ \implies \: \frac{ {(1 - cos \theta)}^{2} }{1 - {cos}^{2} \theta } \\ \\ \implies \: \frac{ {(1 - cos \theta}^{2} }{ {sin}^{2} \theta } \\ \\ \implies \: { \bigg( \frac{1 - cos \theta}{sin \theta} \bigg)}^{2} \\ \\ \implies \: { \bigg(\frac{1}{sin \theta} - \frac{cos \theta}{sin \theta} \bigg)}^{2} \\ \\ \implies \: \: {(cosec \theta - cot \theta)}^{2} \\ \\ \\ hence \: proved \:$