Question

Please answer the question mentioned in the given picture.​​

Answer 1

$\sec + \tan \\ = > \frac{1}{ \cos } + \frac{ \sin}{ \cos} \\ \\ = > \frac{1 + \sin}{ \cos} \\ \\ = > \frac{1 + \sin}{ \cos} \times \frac{1 - \sin }{1 - \sin } \\ \\ = > \frac{1 - \sin {}^{2}}{ \cos(1 - \sin) } \\ \\ = > \frac{ \cos {}^{2}}{ \cos(1 - \sin) } \\ \\ = > \frac{ \cos }{1 - \sin }$

$\huge \mathfrak \red{hence \: proved}$