Question

Prove that :


$\dfrac{ \tan \: A \: + \: \sin \: A }{ \tan A \: - \: \sin \: A } = \: \dfrac{ \sec \:A \: + \: 1}{ \sec \:A \: - \: 1}$
​

Answer 1

Step-by-step explanation:

the required answer is in the attatchment .. and btw sorry for the bad handwriting