Question

If in a triangle ABC ,tanA + tanB + tanC = 6, then find the value of cotA + cotB + cotC . ​

Answer 1

answer :

given tanA +tanB+ tanC =6

• 1/tanA + 1/tanB/ + 1/tanC=1/6
• now WE know that 1/tan = cot so
• cotA+ cotB +cotC = 1/6

Answer 2

Answer:

mark as brilliant

Step-by-step explanation:

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