Hi OldTimer
Been sitting on this for a couple of days, but better late than never.
I'm assuming that you would prefer a leg-up rather than a complete solution, but if that's not the case, or if you need more help, then post again.
For part (i), drop a perpendicular from A onto BC, meeting BC at D say, work out what the angle LAD is in terms of angles of the triangle and then write down a trig identity for this angle.
For part (ii), make the obvious substitutions from part (i) and arrive at (via the sine rule) the requirement that
\(\displaystyle \cot B\cot C+ \cot C \cot A + \cot A \cot B =1.\)
To prove this, make use of the identity
\(\displaystyle \cot (A+B)=\frac{\cot A \cot B - 1}{\cot A + \cot B}.\)
That's not one that you tend to remember but it comes easily from the identity for \(\displaystyle \tan (A+B), \text{(which you should remember)}.\)
(You'll still need to use \(\displaystyle \cot (A+B)=1/\tan(A+B)\)
on the LHS, and remember that A + B + C = 180 deg. )
Tiggsy