Question

prove the following identity
$\frac{ \sin \theta}{1 - \cos \theta} + \frac{ \tan \theta}{1 + \cos \theta} = \sec \theta \cosec \theta + \cot \theta$
​

Answer 1

$\huge \bold {\fbox{ \underline{ \underline \pink{required \: answer}}}}$

we have,

$\bold{ \tt LHS = \frac{ \sin \theta}{1 - \cos \theta} \: + \: \frac{ \tan \theta}{ 1 + \cos \theta} } \\ \\ \bold{ \tt and \: \: \:RHS = \sec \theta \cosec \theta + \cot \theta }$

Solve LHS;

$\bold{ \tt →LHS = \frac{ \sin \theta(1 + \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)} + \frac{ \tan \theta(1 - \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)} } \\ \\ \bold{ \tt →LHS = \frac{ \sin \theta(1 + \cos \theta)}{1 - \cos^{2} \theta} + \frac{ \tan \theta(1 - \cos \theta)}{1 - \cos^{2} \theta } } \\ \\ \bold{ \tt→LHS = \frac{ \sin \theta(1 + \cos \theta)}{ \sin^{2} \theta} + \frac{ \tan \theta(1 - \cos \theta)}{ { \sin}^{2} \theta } } \\ \\ \bold{ \tt →LHS = \frac{ \sin \theta(1 + \cos \theta)}{ { \sin}^{2} \theta } + \frac{ \sin \theta(1 - \cos \theta)}{ \cos \theta \sin^{2} \theta} } \\ \\ \bold{ \tt →LHS = \frac{1 + \cos \theta}{ \sin \theta} + \frac{1 - \cos \theta}{ \cos \theta \sin \theta} } \\ \\ \bold{ \tt →LHS = \frac{1}{ \sin \theta} + \frac{ \cos \theta}{ \sin \theta} + \frac{1}{ \cos \theta \sin \theta} - \frac{ \cos \theta}{ \cos \theta \sin \theta} } \\ \\ \bold{ \tt →LHS = \frac{1}{ \sin \theta} + \frac{ \cos \theta}{ \sin \theta} + \frac{1}{ \cos \theta \sin \theta} - \frac{1}{ \sin \theta} } \\ \\ \bold{ \tt →LHS = \cot \theta + \sec \theta \cosec \theta} \\ \\ \bold{ \implies \:LHS = \sec \theta \cosec \theta + \cot \theta = RHS }$

so, LHS = RHS

$\bold{ \underline{ \underline \orange{hence, \: proved}}}$