Question

prove that root 1 + cos theta by 1 - cos theta is equals to cos theta + cot theta ​

Answer 1

Correct question :-

Prove that ,

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

Proof :-

Formula used :-

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

Explanation :-

Taking left hand side ( LHS)

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

LHS = RHS

hence proved .