TRIGO DOUBT

LHS = cosAsinBsinC+cosBsinCsinA + cosCsinAsinB

= cosAsinA + cosBsinB + cosCsinCsinAsinBsinC

= sin2A + sin2B + sin2C2sinAsinBsinC

= 2sin(A+B)cos(A-B) + 2sinCcosC{cos(A-B) - cos(A+B)}sinC

now since, A+B+C=Ï€,

= 2sinCcos(A-B) + 2sinCcosC{cos(A-B) - cos(A+B)}sinC

= 2sinC{cos(A-B) + cosC}{cos(A-B) - (-cosC)}sinC

= 2 = RHS

Or finish it easily.

Using formula sin2A+sin2B+sin2C=4sinAsinBsinC ,

We get sin2A+sin2B+sin2C2sinAsinBsinC

= 4sinAsinBsinC2sinAsinBsinC

= 2.

Anyways,thank you for the proof.

\sum \frac{\cos A}{\sin B \sin C} = - \sum \frac{\cos (B+C)}{\sin B \sin C}= 3 - \sum \cot A \cot B = 3-1=2

(Since in a triangle ABC, \sum \cot A \cot B = 1)

sir how can u think so nicely!!!

Got the point.

Thank you.