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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Badiuddin