Question

1/sec A -tan A = secA + tanA
​

Answer 1

Hope...

It comforts you...

Answer 2

We have to prove that :

$\frac{1}{ \sec \alpha - \tan\alpha } = \sec \alpha + \tan \alpha$

On taking LHS :

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

=> RHS

HENCE PROVED