Question

If $\frac{\tan A}{1}=\frac{\tan B}{2}=\frac{\tan C}{3}\ne 0$ for a triangle ABC, then

• A. $\tan A+\tan B+\tan C=6$
• B. $\tan A\tan B\tan C=6$
• C. $\tan A\tan B+\tan B\tan C+\tan C\tan A=11$
• D. $\cot A+\cot B+\cot C=6$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\tan A+\tan B+\tan C=6$
B $\tan A\tan B\tan C=6$
C $\tan A\tan B+\tan B\tan C+\tan C\tan A=11$

Let $\frac{\tan A}{1}=\frac{\tan B}{2}=\frac{\tan C}{3}=k$
$\tan A=k$
$\tan B=2k$
$\tan C=3k$
Now,
$\tan A+\tan B+\tan C=\tan A.\tan B.\tan C$
$6k=6k^3$
$k^3-k=0$
$k(k^2-1)=0$
$k=0$ and $k=\pm 1$
Now,
$k=0$ is not possible.
And all the angles cannot be obtuse so $k=-1$ is not possible.
Hence, $k=1$
Therefore, $\tan A=1$ $\tan B=2$ and $\tan C=3$.