proving for tan question

**smplease** · August 11th 2010, 09:45 PM

Hey need some help with this one

Prove that $tan (\theta +\phi ) = (tan \theta + tan \phi) \div ( 1 - tan \theta . tan \phi )$

Cheers

**Amer** · August 11th 2010, 10:21 PM

$\sin ( \theta + \phi ) = \sin \theta \cos \phi + \cos \theta \sin \phi$

$\cos (\theta + \phi ) = \cos \theta \cos \phi - \sin \theta \sin \phi$

$\tan (\theta + \phi ) = \frac{\sin \theta \cos \phi + \cos \theta \sin \phi }{\cos \theta \cos \phi - \sin \theta \sin \phi}$

divide the denominator and nominator by $\cos \phi \cos \theta$

you will get the identity

**smplease** · August 12th 2010, 07:52 PM

thanks for the reply Amer

got the answer out now

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