Dear sameer sai
given
sinA sinB sinC=p
cosA cosB cosC =q
so tanA tanB tanC =p/q
and A+B+C=180
use tan(a+b+c) formula
u will get  tanA+tanB+tanC=tanA tanB tanC
  so        tanA+tanB+tanC=p/q
now
tanAtanB +tanB tanC+ tanC tanA=(sinA sinB cosC +sinB sinC cosA +sinC sinA cosB)/cosA cosB cosC
                                               =1/q (sinA sinB cosC +sinC [sinB cosA + sinA cosB])
                                               =1/q (sinA sinB cosC +sinC [sin(A+B)])
                                               =1/q (sinA sinB cosC +sin2C )
                                               =1/q (sinA sinB cosC +1-cos2C )
                                               =1/q (1+cosC{-cosC+sinA sinB } )
                                               =1/q (1+cosC cosA cosB  )
                                               =1/q (1+q )
now  u can get equation.
            qx3 -px2 +(1+q)x -p =0
 
 

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