Question

The value $\cot x + \cot \left( {{{60}^ \circ } + x} \right) + \cot \left( {{{120}^ \circ } + x} \right)$ is equal to:
A. $\cot 3x$
B. $\tan 3x$
C. $3\tan 3x$
D. $\dfrac{{3 - 9{{\tan }^2}x}}{{3\tan x - {{\tan }^3}x}}$

Answer 1

To solve this question, first we have to write the given expression replacing $\cot x$ with the equivalent terms of $\tan x$. Simplify the equation by substituting the angle values, taking the L.C.M. and writing the formulas of the multiple angles until the desired solution is obtained.

Complete step-by-step solution:
Given expression:
$\cot x + \cot \left( {{{60}^ \circ } + x} \right) + \cot \left( {{{120}^ \circ } + x} \right)$
Now expanding the terms of the simple expressions, we get;
$\Rightarrow \cot x + \dfrac{{\cot 60\cot x - 1}}{{\cot 60 + \cot x}} + \dfrac{{\cot 120\cot x - 1}}{{\cot 120 + \cot x}}$
Replacing the trigonometric value of $\cot x$ in terms of $\tan x$, we get;
$\Rightarrow \dfrac{1}{{\tan x}} + \dfrac{1}{{\tan \left( {{{60}^ \circ } + x} \right)}} + \dfrac{1}{{\tan \left( {{{120}^ \circ } + x} \right)}}$
We have $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
By expanding the obtained equation by substituting it in the above formula, we get;
$\Rightarrow \dfrac{1}{{\tan x}} + \dfrac{1}{{\dfrac{{\tan {{60}^ \circ } + \tan x}}{{1 - \tan {{60}^ \circ }\tan x}}}} + \dfrac{1}{{\dfrac{{\tan {{120}^ \circ } + \tan x}}{{1 - \tan {{120}^ \circ }\tan x}}}}$
We have the standard values of $\tan {60^ \circ }$ and $\tan {120^ \circ }$.
$\tan {60^ \circ } = \sqrt 3$
$\tan {120^ \circ } = - \sqrt 3$
Substituting these values in the above obtained equation, we get;
$\Rightarrow \dfrac{1}{{\tan x}} + \dfrac{1}{{\dfrac{{\sqrt 3 + \tan x}}{{1 - \sqrt 3 \tan x}}}} + \dfrac{1}{{\dfrac{{ - \sqrt 3 + \tan x}}{{1 - \left( { - \sqrt 3 } \right)\tan x}}}}$
Now, simplifying the above equation by sending the denominator of the denominator to the numerator through division, we get;
$\Rightarrow \dfrac{1}{{\tan x}} + \dfrac{{1 - \sqrt 3 \tan x}}{{\tan x + \sqrt 3 }} + \dfrac{{1 + \sqrt 3 \tan x}}{{\tan x - \sqrt 3 }}$
Now, taking the L.C.M. of the two denominators and getting the numerators in the same form by multiplying the denominators of the other fraction, we get;
$\Rightarrow \dfrac{1}{{\tan x}} + \dfrac{{\left( {1 - \sqrt 3 \tan x} \right)\left( {\tan x - \sqrt 3 } \right) + \left( {1 + \sqrt 3 \tan x} \right)\left( {\tan x + \sqrt 3 } \right)}}{{\left( {\tan x + \sqrt 3 } \right)\left( {\tan x - \sqrt 3 } \right)}}$
Now, multiplying the denominators as well as the numerators in respect to the given terms, we get;
$\Rightarrow \dfrac{1}{{\tan x}} + \dfrac{{\tan x - \sqrt 3 {{\tan }^2}x - \sqrt 3 + 3\tan x + \tan x + \sqrt 3 {{\tan }^2}x + \sqrt 3 + 3\tan x}}{{{{\tan }^2}x - 3}}$
Simplifying the terms using simple algebraic expressions and methods, we get;
$\Rightarrow \dfrac{1}{{\tan x}} + \dfrac{{2\tan x + 6\tan x}}{{{{\tan }^2}x - 3}}$
Adding the numerator with like terms, we get;
$\Rightarrow \dfrac{1}{{\tan x}} + \dfrac{{8\tan x}}{{{{\tan }^2}x - 3}}$
Now taking the L.C.M. of the denominators and multiplying the numerators with the opposite denominator, we get;
$\Rightarrow \dfrac{{{{\tan }^2}x - 3 + 8{{\tan }^2}x}}{{{{\tan }^3}x - 3\tan x}}$
Simplifying the numerator, we get;
$\Rightarrow \dfrac{{9{{\tan }^2}x - 3}}{{{{\tan }^3}x - 3\tan x}}$
Taking the negative sign out from both the numerator and the denominator, we get;
$\Rightarrow \dfrac{{3 - 9{{\tan }^2}x}}{{3\tan x - {{\tan }^3}x}}$
Taking the common terms in the numerator and denominator out, we get;
$\Rightarrow \dfrac{{ - 3\left( {1 - 3{{\tan }^2}x} \right)}}{{ - \left( {3\tan x - {{\tan }^2}x} \right)}}$
Cancelling out the negative sign, we get;
$\Rightarrow \dfrac{{3\left( {1 - 3{{\tan }^2}x} \right)}}{{\left( {3\tan x - {{\tan }^2}x} \right)}}$
Bringing the numerator trigonometric part to the denominator for better access, we get;
$\Rightarrow \dfrac{3}{{\left( {\dfrac{{3\tan x - {{\tan }^2}x}}{{1 - 3{{\tan }^2}x}}} \right)}}$
Now, the denominator is in the form of $\tan 3x$
$\tan 3x = \dfrac{{3\tan x - {{\tan }^2}x}}{{1 - 3{{\tan }^2}x}}$
Substituting the denominator with the above expression, we get;
$\Rightarrow \dfrac{3}{{\tan 3x}}$
Now, substituting the $\tan x$ in terms of $\cot x$, we get;
$\Rightarrow 3\cot 3x$
Since we do not have the final answer in the options, we have the mid value answer which is,
$\Rightarrow \dfrac{{3 - 9{{\tan }^2}x}}{{3\tan x - {{\tan }^3}x}}$
Therefore, for the given expression, we have;
$\cot x + \cot \left( {{{60}^ \circ } + x} \right) + \cot \left( {{{120}^ \circ } + x} \right) = \dfrac{{3 - 9{{\tan }^2}x}}{{3\tan x - {{\tan }^3}x}}$

$\therefore$ The correct option is D.

Note: We have to know that trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. The Greeks focused on the calculations of chords, while mathematics in India created the earliest-known tables of values for trigonometric ratios (also called trigonometric functions) such a sine.