Hey need some help with this one
Prove that $\displaystyle tan (\theta +\phi ) = (tan \theta + tan \phi) \div ( 1 - tan \theta . tan \phi )$
Cheers
$\displaystyle \sin ( \theta + \phi ) = \sin \theta \cos \phi + \cos \theta \sin \phi $
$\displaystyle \cos (\theta + \phi ) = \cos \theta \cos \phi - \sin \theta \sin \phi $
$\displaystyle \tan (\theta + \phi ) = \frac{\sin \theta \cos \phi + \cos \theta \sin \phi }{\cos \theta \cos \phi - \sin \theta \sin \phi} $
divide the denominator and nominator by $\displaystyle \cos \phi \cos \theta $
you will get the identity