A geometry problem by Nivedit Jain

$\begin{aligned} X & = \frac {\cos A}{\sin B \sin C} + \frac {\cos B}{\sin C \sin A} + \frac {\cos C}{\sin A \sin B} & \small \color{#3D99F6} \text{For }\triangle, \ A = 180^\circ - B - C \\ & = \frac {\cos (180^\circ - B-C)}{\sin B \sin C} + \frac {\cos (180^\circ -C-A)}{\sin C \sin A} + \frac {\cos (180^\circ -A- B)}{\sin A \sin B} & \small \color{#3D99F6} \text{Note: } \cos (180^\circ - \theta) = - \cos \theta \\ & = - \frac {\cos (B+C)}{\sin B \sin C} - \frac {\cos (C+A)}{\sin C \sin A} - \frac {\cos (A+B)}{\sin A \sin B} \\ & = \frac {\sin B \sin C - \cos B \cos C}{\sin B \sin C} + \frac {\sin C \sin A - \cos C \cos A}{\sin C \sin A} + \frac {\sin A \sin B - \cos A \cos B}{\sin A \sin B} \\ & = 3 - ({\color{#3D99F6}\cot B \cot C + \cot C \cot A + \cot A \cot B}) & \small \color{#3D99F6} \text{See note: }\sum_{cyc} \cot A \cot B = 1 \\ & = 3 - {\color{#3D99F6}1} \\ & = \boxed{2} \end{aligned}$

---

Note: For triangle $ABC$ ,

$\small \begin{aligned} \tan A + \tan B + \tan C & = \tan A \tan B \tan C \\ \cot B \cot C + \cot C \cot A + \cot A \cot B & = 1 \end{aligned}$

0.5cosecAcosecBcosecC[sin2A+sin2B+sin2c] taking LCM Now 0.5cosecAcosecBcosecC[2sin(A+C)cos(A-C) +2sinBcosB] Sin(A+C)= Sin(180° - B)=sinB taking 2sinB common and cancelling we get cosecAcosecC[cos(A-C) + cos(B)] =cosecAcosecC[ 2cos((C+B-A)/2)cos((A+B-C)/2)] C+B=180°-A A+B=180°-C We get 2cosecAcosecCsinAsinB We get 2 that is the answer ::)