Question

If $tanA+tanB+tanC=6$ and $tanA.tanB=3$, then $tanC$ is equal to

• A. 2
• B. 1
• C. 3
• D. -1

Answer 1

The correct option is A 2

In a triangle,

$A+B+C=180°$ (Where A,B and C are angles of triangle)

Hence, $(A+B)=180°-C$ ---------------- (1)

Now given that,

$tanA+tanB+tanC=6$

$tanA+tanB=6-tanC$

Dividing the whole equation by $(1-tanA.tanB)$,

$\frac{tanA+tanB}{1-tanA.tanB}=\frac{6-tanC}{1-tanA.tanB}$

$tan(A+B)=\frac{6-tanC}{1-tanA.tanB}$ $\left[tan\right(A+B)=\frac{tanA+tanB}{1-tanA.tanB}]$

$tan(180°-C)=\frac{6-tanC}{1-tanA.tanB}$ [from (1)]

$-tanC=\frac{6-tanC}{1-tanA.tanB}$ $\left[tan\right(180°-C)=-tanC]$

Given, $tanA.tanB=3$. So,

$-tanC=\frac{6-tanC}{1-3}$

$(-2)\times (-tanC)=6-tanC$

$2tanC+tanC=6$

$3tanC=6$

Hence,

$tanC=2$