Question

tan theta plus sin theta by tan theta minus sin theta equal to sec theta plus 1 by sec theta minus 1

Answer 1

just divide by sin theta

Answer 2

$\frac{ \tan \theta + \sin \theta }{\tan \theta - \sin \theta } = \frac{ \sec \theta + 1}{ \sec \theta - 1}$
L.H.S.:
$= \frac{ \frac{ \sin \theta }{ \cos \theta} + \sin \theta }{ \frac{ \sin \theta }{ \cos \theta} - \sin \theta } \\ = \frac{ \frac{ \sin \theta + \sin \theta .\cos \theta}{ \cos \theta} }{ \frac{ \sin \theta - \sin \theta .\cos \theta}{ \cos \theta}} \\ = \frac{ \sin \theta(1 + \cos \theta)}{\sin \theta(1 - \cos \theta)} \\ = \frac{1 + \frac{1}{ \sec \theta} }{1 - \frac{1}{ \sec \theta} } \\ = \frac{ \frac{ \sec \theta + 1}{ \sec \theta} }{\frac{ \sec \theta - 1}{ \sec \theta} } \\ = \frac{ \sec \theta + 1}{ \sec \theta - 1}$
= R.H.S.

Hence, proved.

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