Question

(cosec#-cot#2)^2=1-cos#/1+cos#

Answer 1

Answer:

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Answer 2

AnswEr:-

$\bf{ (i) {(cosecθ - cotθ)}^{2} } \\ \bf = { { (\frac{1}{sinθ} - \frac{cosθ}{sinθ}) }^{2} } \\ = \bf{ { (\frac{1 - cosθ}{sinθ}) }^{2} } \\ = \bf{ \frac{ {(1 - cosθ)}^{2} }{ {sin}^{2}θ } } \\ = \bf{ \frac{(1 - cosθ)(1 - cosθ)}{1 - {cos}^{2}θ } } \\ = \bf{\frac{(1 - cosθ)(1 - cosθ)}{(1 - cosθ) (1 + cosθ)} } \\ = \bf{ \frac{1 - cosθ}{1 + cosθ}} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge{ \underline{ \blue{ \underline{ \mathcal{Proved}}}}}$