Question
In a $$\vartriangle ABC,\tan A \cdot \tan B \cdot \tan C = 9.$$ For such triangles, if $${\tan ^2}A + {\tan ^2}B + {\tan ^2}C = k$$ then
A.
$$9 \cdot \root 3 \of 3 < k < 27$$
B.
$$k \leqslant 27$$
C.
$$k < 9 \cdot \root 3 \of 3 $$
D.
$$k < 27$$
Answer :
$$k \leqslant 27$$
Solution :
$$\eqalign{
& 3\left( {{{\tan }^2}A + {{\tan }^2}B + {{\tan }^2}C} \right) - {\left( {\tan A + \tan B + \tan C} \right)^2} = {\left( {\tan A - \tan B} \right)^2} + {\left( {\tan B - \tan C} \right)^2} + {\left( {\tan C - \tan A} \right)^2} > 0\,\,{\text{for here }}\tan A = \tan B = \tan C = \sqrt 3 \,\,{\text{is not true}} \cr
& {\text{or, }}3k - {\left( {\tan A \cdot \tan B \cdot \tan C} \right)^2} > 0\,\,{\text{because in the }}\vartriangle ABC,\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C \cr
& {\text{or, }}3k - 81 > 0\,\,\,{\text{or, }}k > 27. \cr} $$