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Let A,B,C be angles of triangles with vertex A -= (4,-1) and internal angular bisectors of angles B and C be x - 1 = 0 and x - y - 1 = 0 respectively. If A,B,C are angles of triangle at vertices A,B,C respectively then cot ((B)/(2))cot .((C)/(2)) =

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Solution :Angle between `X -1 =0` and BC is `(B)/(2) rArr TAN.(B)/(2) =(1)/(2)`
Angle between `x -y -1 =0` and BC is `(C )/(2) rArr tan.(C )/(2) =(1)/(3)`
`rArr cot.(B)/(2) cot.(C )/(2) =6`


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