Question

Prove the above question​

Answer 1

$lhs = \frac{sin \alpha + tan \alpha }{cos \alpha } \\ \\ = \frac{sin \alpha + \frac{sin \alpha }{cos \alpha } }{cos \alpha } \\ \\ = \frac{ \frac{sin \alpha (cos \alpha ) + sin \alpha }{cos \alpha } }{cos \alpha } \\ \\ = \frac{sin \alpha (cos \alpha + 1)}{ {cos}^{2} \alpha } \\ \\ = \frac{tan \alpha (cos \alpha + 1)}{cos \alpha } \\ \\ = \frac{tan \alpha. cos \alpha + tan \alpha }{cos \alpha } \\ \\ = \frac{tan \alpha .cos \alpha }{cos \alpha } + \frac{tan \alpha }{cos \alpha } \\ \\ = tan \alpha \: + \frac{tan \alpha }{cos \alpha } \\ \\ = tan \alpha (1 + \frac{1}{cos \alpha } ) \\ \\ = tan \alpha (1 + sec \alpha ) = rhs$

mark as brainliest.......✌✌

Answer 2

Answer:

hope it helps u

have a nice day