Prove that tan(45 + A)........

**milan** · March 3rd 2008, 12:00 PM

Prove that tan(45+A) = cosA + sinA/cosA - sinA

I converted tan(45+A) into 1+tanA/1 - tanA

but i dont know what to do next? Im sure its just a small thing but i just cant get it! Some help please!

**bobak** · March 3rd 2008, 12:23 PM

$\tan \left( \frac{\pi}{4} + A \right) = \frac{ \tan( \frac{\pi}{4} ) + \tan A}{1 - \tan( \frac{\pi}{4}) \tan A }$

$\Rightarrow \ \ \frac{ 1 + \tan A}{1 - \tan A }$

$\Rightarrow \ \ \frac{ 1 + \frac{\sin A}{\cos A}}{1 - \frac{\sin A}{\cos A} } \ \ = \ \ \frac{ 1 + \frac{\sin A}{\cos A}}{1 - \frac{\sin A}{\cos A} } \times \frac{\cos A }{\cos A }$

$\Rightarrow \frac{ \cos A + \sin A}{\cos A - \sin A}$

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