Question

prove that (cot theta -cos theta) / ( cot theta + cos theta)= (cosec theta -1)/(cosec theta +1).
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Answer 1

To Prove :

$\longmapsto \sf \frac{ \cot \theta - \cos\theta}{\cot\theta + \cos\theta} = \frac{ \csc\theta - 1}{ \csc\theta + 1}$

Solution :

On solving LHS

$\longmapsto \sf\frac{ \cot\theta - \cos\theta}{\cot\theta + \cos\theta}$

$\longmapsto \sf\dfrac{ \dfrac{ \cos\theta}{ \sin\theta} - \cos\theta}{\dfrac{ \cos\theta}{ \sin\theta}+ \cos\theta}$

$\longmapsto \sf\dfrac{ \dfrac{ \cos \theta- \sin\theta \cos\theta}{ \cancel{\sin\theta}}}{\dfrac{ \cos \theta+ \sin\theta \cos\theta}{ \cancel{\sin\theta}}}$

$\longmapsto \sf\dfrac{ \cos\theta - \sin\theta \cos\theta}{ \cos\theta + \sin\theta \cos\theta}$

$\longmapsto \sf \dfrac{\cancel{ \cos\theta} (1- \sin \theta)}{ \cancel{\cos\theta}(1 + \sin\theta)}$

$\longmapsto \sf\dfrac{ 1 - \sin\theta}{ 1+ \sin\theta}$

$\longmapsto \sf\dfrac{ 1 - \dfrac{1}{ \csc\theta} }{ 1+ \dfrac{1}{ \csc\theta} }$

$\longmapsto \sf\dfrac{ \dfrac{ \csc\theta - 1}{ \cancel{ \csc\theta}} }{ \dfrac{ \csc\theta + 1}{ \cancel{\csc\theta}} }$

$\longmapsto \sf\frac{ \csc \theta- 1}{ \csc \theta+ 1}$

LHS = RHS

Hence proved