Question:
If in a triangle ABC , tanA tanB tanC=6,then cotAcotBcotC=
Answer:
Given tanA + tanB + tanC = 6
Now tan(A + B + C) = {(tanA + tanB + tanC) - tanA*tanB*tanC}/{1 - (tanA*tanB + tanB*tanC + tanA*tanC)}
We know that,
A + B + C = π
=> tan(A + B + C) = tan π
=> => tan(A + B + C) = 0
Now
0 = {(tanA + tanB + tanC) - tanA*tanB*tanC}/{1 - (tanA*tanB + tanB*tanC + tanA*tanC)}
=> tanA + tanB + tanC - tanA*tanB*tanC = 0
=> tanA + tanB + tanC = tanA*tanB*tanC
=> tanA*tanB*tanC = 6
=> (1/cotA)*(1/cotB)*(1/cotC) = 6
=> 1/(cotA*cotB*cotC) = 6
=> cotA*cotB*cotC = 1/6
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