In a triangle ABC, if sides a,b and c are in A.P(Arithmetic Progression), angle B = pi/4, find tan A/2 + tan C/2?
This is how I would start off.
$\displaystyle \tan \frac{A}{2} + \tan \frac{C}{2} = \frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} + \frac{\sin \frac{C}{2}}{\cos \frac{C}{2}} = \frac{\sin \left( \frac{A}{2} + \frac{C}{2} \right)}{\cos \frac{A}{2}\cos \frac{C}{2}}$
But $\displaystyle A+B+C = \pi$ and $\displaystyle B=\frac{\pi}{4}$ this means $\displaystyle \sin \left( \frac{A}{2}+\frac{C}{2} \right) = \sin \frac{3\pi}{8}$.
Thus, the problem comes down to computing $\displaystyle \cos \frac{A}{2}\cos \frac{C}{2}$.
Which looks easier.