1. Find A

Show that $secA+tanA=tan(\frac{\pi}{4}+\frac{A}{2})$ and deduce a similar expression for $secA-tanA$
Hence find the surd form of the values of $tan\frac{7\pi}{12}$ and $tan\frac{\pi}{12}$
i have completed the first two parts, for the second i got the equality $tan(\frac{\pi}{4}-\frac{A}{2})$
now i only have the last part to do and i'm lost.

2. I am not completely sure about using $\tan(\frac{\pi}{4}-\frac{A}{2})$ to get to the surd value for $tan\frac{7\pi}{12}$ and $tan\frac{\pi}{12}$, however I do know another way:

You can try represent $\frac{7\pi}{12}$ and $\frac{1\pi}{12}$ with angles that we now the surd form:
$\frac{7\pi}{12} = \frac{4\pi}{12} + \frac{3\pi}{12}$
$\frac{7\pi}{12} = \frac{4\pi}{12} - \frac{3\pi}{12}$
By using a double angle formula, you can now find the exact value of $tan\frac{7\pi}{12}$ and $tan\frac{\pi}{12}$.

3. Originally Posted by arze
Show that $secA+tanA=tan(\frac{\pi}{4}+\frac{A}{2})$ and deduce a similar expression for $secA-tanA$
Hence find the surd form of the values of $tan\frac{7\pi}{12}$ and $tan\frac{\pi}{12}$
i have completed the first two parts, for the second i got the equality $tan(\frac{\pi}{4}-\frac{A}{2})$
now i only have the last part to do and i'm lost.
You require $\frac{\pi}{4} + \frac{A}{2} = \frac{7 \pi}{12} \Rightarrow A = \frac{2 \pi}{3}$. So substitute $A = \frac{2 \pi}{3}$ into the first expression.

Similarly, substitute $A = \frac{\pi}{3}$ into the second expression.