Question

If A, B, and C are acute angles such that $\tan A + \tan B + \tan C = \tan A\tan B\tan C$. Then $\cot A\cot B\cot C = $
a) $ \leqslant \dfrac{1}{{\sqrt 3 }}$
b) $ \leqslant \dfrac{1}{{2\sqrt 3 }}$
c) $ \leqslant \dfrac{1}{{3\sqrt 3 }}$
d)None of these

Answer 1

Hint: Acute angles are the angle which is less than ${90^0}$. We always know that if the angles are acute angles the Arithmetic mean of three angles is greater than the geometric mean of the three angles. Arithmetic mean is nothing but the average of three angles and the geometric mean is the cube root of three angles.
Formula:
The Arithmetic mean of three angles is less than the geometric mean of the three angles. Since,
$Arithmetic\ mean \geqslant geometric\ mean$
The Arithmetic mean is nothing but the average of three values.
$Arithmetic\ mean = \dfrac{{a + b + c}}{3}$
The geometric mean is the cube root of three values.
$geometric\ mean = \sqrt[3]{{a \times b \times c}}$

Complete step-by-step answer:
Given,
Three angles are given such as $A$,$B$and $C$.
The given condition $\tan A + \tan B + \tan C = \tan A\tan B\tan C$
We have to find $\cot A + \cot B + \cot C = $
We know the condition that if the angles are acute angle the Arithmetic mean of three angles are greater than the geometric mean of the three angles.
According to the question,
Substitute $a = \tan A$, $b = \tan B$and $c = \tan C$.
We know that the arithmetic mean $Arithmetic\ mean = \dfrac{{a + b + c}}{3}$
We know that the geometric mean $geometric\ mean = \sqrt[3]{{a \times b \times c}}$
Substitute $a = \tan A$, $b = \tan B$and $c = \tan C$.
$Arithmetic\ mean = \dfrac{{\tan A + \tan B + \tan C}}{3}$
$geometric\ mean = \sqrt[3]{{\tan A \times \tan B \times \tan C}}$
We know that the relation between the arithmetic mean and geometric mean,
$Arithmetic\ mean \geqslant geometric\ mean$
Substituting $Arithmetic\ mean = \dfrac{{\tan A + \tan B + \tan C}}{3}$ and $geometric\ mean = \sqrt[3]{{\tan A \times \tan B \times \tan C}}$
$\dfrac{{\tan A + \tan B + \tan C}}{3} \geqslant \sqrt[3]{{\tan A \times \tan B \times \tan C}}$
According to the given, $\tan A + \tan B + \tan C = \tan A\tan B\tan C$
Converting the cube root to the power of another side,
$\dfrac{{{{\left( {\tan A + \tan B + \tan C} \right)}^3}}}{{{3^3}}} \geqslant \tan A \times \tan B \times \tan C$
Expanding cube value in the denominator,
$\dfrac{{{{\left( {\tan A + \tan B + \tan C} \right)}^3}}}{{27}} \geqslant \tan A \times \tan B \times \tan C$
By substituting $\tan A + \tan B + \tan C = \tan A\tan B\tan C$, we get
$\dfrac{{{{\left( {\tan A\tan B\tan C} \right)}^3}}}{{27}} \geqslant \tan A \times \tan B \times \tan C$
Bringing terms from right side to left side, we get
$\dfrac{{{{\left( {\tan A\tan B\tan C} \right)}^3}}}{{27 \times \tan A \times \tan B \times \tan C}} \geqslant 1$
Dividing the terms between the numerator and the equator, we get,
$\dfrac{{{{\left( {\tan A\tan B\tan C} \right)}^2}}}{{27}} \geqslant 1$
By reciprocating the terms, we get
$\dfrac{{27}}{{{{\left( {\tan A\tan B\tan C} \right)}^2}}} \leqslant 1$
We know $\cot A = \dfrac{1}{{\tan A}}$, by substituting we get
${\left( {\cot A\cot B\cot C} \right)^2} \times 27 \leqslant 1$
By bringing the value in the numerator to the denominator of another side, we get,
${\left( {\cot A\cot B\cot C} \right)^2} \leqslant \dfrac{1}{{27}}$
Squaring on both sides we get,
$\left( {\cot A\cot B\cot C} \right) \leqslant \dfrac{1}{{3\sqrt 3 }}$

So, the correct answer is “Option C”.

Note: Always remember to find the relation between the arithmetic mean and geometric mean which is if the angles are acute angle the Arithmetic mean of three angles are greater than the geometric mean of the three angles. Remember tan is reciprocal to the cot. If we reciprocate the terms the lesser sign will be changed to a greater sign.