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

Hint: 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.