Question

If \[A,{\rm{ }}B,{\rm{ }}C\] are acute angles of a triangle such that \[{\rm{\tan}}A{\rm{ }} + {\rm{ \tan}}B{\rm{ }} + {\rm{ \tan}}C{\rm{ }} = {\rm{ \tan}}A{\rm{ \tan}}B{\rm{ \tan}}C\]. Then find the value of \[\cot A{\rm{ cot}}B{\rm{ cot}}C\].
A. \[ \le \dfrac{1}{{\surd 3}}{\rm{ }}\]
B. \[ \le \dfrac{1}{{2\surd 3}}{\rm{ }}\]
C. \[ \le \dfrac{1}{{3\surd 3}}{\rm{ }}\]
D. none of these

Answer 1

Hint: we will use the result between arithmetic mean (AM) and geometric mean (GM). Which says that the AM shall Always greater than or equals to the GM. Then we will substitute the given values in the results and equate them. Finally obtaining the required value.

Formula used:

\[\text{Arithmetic mean (AM)} \ge \text{geometric mean (GM)} \] that is the Arithmetic mean will always be greater than or equals to the geometric mean.
\[\left[ {\dfrac{{{\rm{ }}A + B + C}}{3}} \right]{\rm{ }} \ge \sqrt[3]{{A{\rm{ }}B{\rm{ }}C}}\]
Where \[\dfrac{{{\rm{ }}A + B + C}}{3}\] is the Arithmetic mean (AM) of three terms and \[\sqrt[3]{{A{\rm{ }}B{\rm{ }}C}}\] is the geometric mean (GM) of term \[A,{\rm{ }}B,{\rm{ }}C\]. where\[A,{\rm{ }}B,{\rm{ }}C\] can be the variable or the cons\tant as per question.

Complete step by step solution:

We are given that \[A,{\rm{ }}B,{\rm{ }}C\] are acute angles of a triangle such that \[{\rm{\tan}}A{\rm{ }} + {\rm{ \tan}}B{\rm{ }} + {\rm{ \tan}}C{\rm{ }} = {\rm{ \tan}}A{\rm{ \tan}}B{\rm{ \tan}}C\].
The result with which we will procced here is \[\left[ {\dfrac{{{\rm{ }}A + B + C}}{3}} \right]{\rm{ }} \ge \sqrt[3]{{A{\rm{ }}B{\rm{ }}C}}\]
Substitute the given values \[{\rm{\tan }}A,{\rm{ \tan }}B,{\rm{ \tan }}C\] in place of \[A,{\rm{ }}B,{\rm{ }}C\] in the result:
\[\left[ {\dfrac{{\tan A{\rm{ }} + {\rm{ \tan }}B{\rm{ }} + {\rm{ \tan }}C}}{3}} \right]{\rm{ }} \ge \sqrt[3]{{{\rm{\tan }}A{\rm{ \tan }}B{\rm{ \tan }}C}}{\rm{ }}\]
Rearrange the terms to evaluate the expression,
\[\left[ {\dfrac{{\tan A{\rm{ }} + {\rm{ \tan }}B{\rm{ }} + {\rm{ \tan }}C}}{3}} \right]{\rm{ }} \ge {\rm{ }}{\left( {{\rm{\tan }}A{\rm{ \tan }}B{\rm{ \tan }}C} \right)^{\dfrac{1}{3}}}\]
Cross multiply and then cube the terms on the left-hand side and right-hand side then we get,
\[{\left[ {{\rm{\tan }}A{\rm{ }} + {\rm{ }}t{\rm{an }}B{\rm{ }} + {\rm{ \tan }}C} \right]^3}\; \ge {\rm{ }}27{\rm{ }}\left( {{\rm{\tan }}A{\rm{ \tan }}B{\rm{ \tan }}C} \right)\]
Substitute the value of left-hand side as \[{\rm{\tan}}A{\rm{ }} + {\rm{ \tan}}B{\rm{ }} + {\rm{ \tan}}C{\rm{ }} = {\rm{ \tan}}A{\rm{ \tan}}B{\rm{ \tan}}C\]:
\[{\left( {{\rm{\tan }}A{\rm{ \tan }}B{\rm{ \tan }}C} \right)^3}\; \ge {\rm{ }}27{\rm{ }}\left( {{\rm{\tan }}A{\rm{ \tan }}B{\rm{ \tan }}C} \right){\rm{ }}\]
Canceling the common term and then square root on the left-hand side and right-hand side to get,
\[\left( {{\rm{\tan }}A{\rm{ \tan }}B{\rm{ \tan }}C} \right)\; \ge {\rm{ }}3\sqrt 3 \]
Reciprocate the terms on the left-hand side and right-hand side to get the required solution.
\[\dfrac{1}{{\left( {\tan{\rm{ }}A{\rm{ }}\tan{\rm{ }}B{\rm{ }}\tan{\rm{ }}C} \right)}} \le \dfrac{1}{{3\surd 3}}{\rm{ }}\]
This implies that,
\[{\rm{cot }}A{\rm{ cot }}B{\rm{ cot }}C{\rm{ }} \le \dfrac{1}{{3\surd 3}}{\rm{ }}\]

Hence option C is correct.

Note: This problem can also be solved by cubing the given terms with the help of a formula. One should take great care while cube rooting terms on left-hand side and right-hand side also alertness is required while reciprocating the terms.