+118687 I expect a number of people, including me, are struggling with your question OldTimer.
+118687 Hi Oldtimer,
Isn't it great when you can work these questons out yourself! 
I played with it but I thought that L M and N were points on the triangle. Not on the circle.
It says on opposite sides, to me it made more sense to think that the triangle was meant.
So i guess I had no hope from the outset. 
+397 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
+26393 O is the centre and R the radius of the circle circumscribing the triangle ABC, AO, BO, and CO meet the opposite sides in L, M, N respectively.
Show that
\((i)\quad AL = \dfrac{ b\sin C} { \cos(B-C) }\)
\(\text{Let $\angle BAO= \angle ABO = w$ } \\ \text{Let $\angle CAO= \angle ACO = v$ } \\ \text{Let $\angle CBO= \angle BCO = u$ } \\ \text{Let $\angle A= v+w$ } \\ \text{Let $\angle B= w+u$ } \\ \text{Let $\angle C= u+v$ } \\ \text{Let $\angle CLA = \varphi$} \)
\(\begin{array}{|rcll|} \hline A+B-C &=& (v+w)+(w+u)-(u+v) \\ A+B-C &=& 2w \quad | \quad A+B = 180^\circ - C \\ 180^\circ - C-C &=& 2w \\ 180^\circ -2C &=& 2w \quad | \quad : 2 \\ 90^\circ -C &=& w \\ \mathbf{w} &=& \mathbf{90^\circ -C} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline A-B+C &=& (v+w)-(w+u)+(u+v) \\ A-B+C &=& 2v \quad | \quad A+C = 180^\circ - B \\ 180^\circ - B-B &=& 2v \\ 180^\circ -2B &=& 2v \quad | \quad : 2 \\ 90^\circ -B &=& v \\ \mathbf{v} &=& \mathbf{90^\circ -B} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline -A+B+C &=& -(v+w)+(w+u)+(u+v) \\ -A+B+C &=& 2u \quad | \quad B+C = 180^\circ - A \\ 180^\circ - A-A &=& 2u \\ 180^\circ -2A &=& 2u \quad | \quad : 2 \\ 90^\circ -A &=& u \\ \mathbf{u} &=& \mathbf{90^\circ -A} \\ \hline \end{array}\)
In triangle ALC:
\(\begin{array}{|rcll|} \hline \varphi &=& 180^\circ-(v+C) \quad | \quad \mathbf{v=90^\circ -B} \\ \varphi &=& 180^\circ-(90^\circ -B+C)\\ \varphi &=& 180^\circ-90^\circ -(-B+C)\\ \varphi &=& 180^\circ-90^\circ -(C-B)\\ \varphi &=& 90^\circ -(C-B)\\ \mathbf{\varphi} &=& \mathbf{90^\circ +(B-C)} \\ \hline \end{array}\)
In triangle ALC - sin rule:
\(\begin{array}{|rcll|} \hline \dfrac{ \sin{C} } {AL} &=& \dfrac{\sin(\varphi)}{b} \\\\ \dfrac{ \sin{C} } {AL} &=& \dfrac{\sin\Big(90^\circ +(B-C)\Big)}{b} \\\\ \dfrac{ \sin{C} } {AL} &=& \dfrac{\cos\Big(B-C\Big)}{b} \quad | \quad \updownarrow \\\\ \dfrac{AL}{ \sin{C} } &=& \dfrac{b} {\cos\Big(B-C\Big)}\\\\ \mathbf{AL} &=& \mathbf{\dfrac{b\sin{C}} {\cos\Big(B-C\Big)}} \\ \hline \end{array}\)
