The product of the sines of the angles of a triangle is `p` and the product of their cosines is `qdot` Show that the tangents of the angles are the roots of the equation `q x^3-p x^2+(1+q)x-p=0.`

Explanation

From the question.
`sin A sin B sin C=p` and `cos A cos B cos C=q`
`therefore tan A tan B tan C=p//q`
Also, `tan A+tan B+tan C=tan A tan B tanC=p//q`
Now, `tan A tan B +tan B tan C+tan C tan A`
`=(sin A sin B cos C+sin B sinC cos A+sin C sin a cos B)/(cos A cos B cos C)`
`=(1)/(2q)[(sin^(2)A+sin^(2)B-sin^(2)C)+(sin^(2)N+sin^(2)C-sin^(2)A)+sin^(2)C+sin^(2)A-sin^(2)B)]`
`[because A+B+C=pi "and" 2 sin A sin B cos C=sin^(2)A+sin^(2)B-sin^(2)C]`
`=(1)/(2q)[sin^(2)A+sin^(2)B+sin^(2)C]`
`=(1)/(4q)[3-(cos 2A+cos2B+cos2C)]`
`=(1)/(q)[1+cosA cos B cos C]=(1)/(q)(1+q)`
The equation whose roots are `tan A,tan B` and `tan C` will be given by `x^(3)-(tan A+tanB+tan C)x^(2)+(tan A tan B+tan B tan C+tan Ctan A)x-tanA tan B tan C=0`
or `x^(3)-(p)/(q)x^(2)+(1+q)/(q)x-(p)/(q)=0`
or `qx^(3)-px^(2)+(1+q)x-p=0`