Question

plz answer the questions that are ticked marked, plz plz plz.!!​

Answer 1

Question :-

Prove that ,

$\mathfrak{ \sqrt{ \frac{1 - \cos \: \theta}{1 + \cos \: \theta} } = \csc \: \theta \: - \cot \: \theta}$

Formula used :-

$\star \: 1 - { \cos}^{2} \theta = { \sin}^{2} \theta \: \\ \\ \star \: \frac{1}{ \sin \: \theta} = \csc \: \theta \\ \\ \star \: \frac{ \cos \: \theta}{ \sin \: \theta} = \cot \: \theta$

Solution :-

Taking Left hand side (LHS)

$\rightarrow \: \sqrt{ \frac{1 - \cos \: \theta}{1 + \cos \: \theta} } \: \\ \\ \sf{rationalising \: by \: \sqrt{1 - \cos \: \theta} } \\ \\ \rightarrow \: \sqrt{ \frac{1 - \cos \: \theta}{1 + \cos \: \theta} } \times \sqrt{ \frac{1 - \cos \: \theta}{1 - \cos \: \theta} } \\ \\ \rightarrow \: \sqrt{ \frac{ {(1 - \cos \: \theta)}^{2} }{(1 + \cos \: \theta)(1 - \cos \: \theta)} } \\ \\ \rightarrow \: \sqrt{ \frac{ {(1 - \cos \: \theta)}^{2} }{1 - { \cos}^{2} \theta} } \\ \\ \rightarrow \: \sqrt{ \frac{ {(1 - \cos \: \theta)}^{2} }{ { \sin}^{2} \theta \: } } \\ \\ \rightarrow \: \sqrt{ { \bigg( \frac{1 - \cos \: \theta}{ \sin \: \theta} \bigg)}^{2} } \\ \\ \rightarrow \: \frac{1}{ \sin \: \theta} - \frac{ \cos \: \theta}{ \sin \: \theta} \\ \\ \rightarrow \: \csc \: \theta \: - \cot \: \theta \:$

LHS = RHS

hence prove .

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