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. $${\displaystyle \frac{3-9{\tan }^{2}x}{3\tan x-{\tan }^{3}x}}$$

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;
$$\Rightarrow \cot x+{\displaystyle \frac{\cot 60\cot x-1}{\cot 60+\cot x}}+{\displaystyle \frac{\cot 120\cot x-1}{\cot 120+\cot x}}$$
Replacing the trigonometric value of $$\cot x$$ in terms of $$\tan x$$, we get;
$$\Rightarrow {\displaystyle \frac{1}{\tan x}}+{\displaystyle \frac{1}{\tan \left({60}^{\circ }+x\right)}}+{\displaystyle \frac{1}{\tan \left({120}^{\circ }+x\right)}}$$
We have $$\tan \left(A+B\right)={\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B}}$$
By expanding the obtained equation by substituting it in the above formula, we get;
$$\Rightarrow {\displaystyle \frac{1}{\tan x}}+{\displaystyle \frac{1}{{\displaystyle \frac{\tan {60}^{\circ }+\tan x}{1-\tan {60}^{\circ }\tan x}}}}+{\displaystyle \frac{1}{{\displaystyle \frac{\tan {120}^{\circ }+\tan x}{1-\tan {120}^{\circ }\tan x}}}}$$
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;
$$\Rightarrow {\displaystyle \frac{1}{\tan x}}+{\displaystyle \frac{1}{{\displaystyle \frac{\sqrt{3}+\tan x}{1-\sqrt{3}\tan x}}}}+{\displaystyle \frac{1}{{\displaystyle \frac{-\sqrt{3}+\tan x}{1-\left(-\sqrt{3}\right)\tan x}}}}$$
Now, simplifying the above equation by sending the denominator of the denominator to the numerator through division, we get;
$$\Rightarrow {\displaystyle \frac{1}{\tan x}}+{\displaystyle \frac{1-\sqrt{3}\tan x}{\tan x+\sqrt{3}}}+{\displaystyle \frac{1+\sqrt{3}\tan x}{\tan x-\sqrt{3}}}$$
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;
$$\Rightarrow {\displaystyle \frac{1}{\tan x}}+{\displaystyle \frac{\left(1-\sqrt{3}\tan x\right)\left(\tan x-\sqrt{3}\right)+\left(1+\sqrt{3}\tan x\right)\left(\tan x+\sqrt{3}\right)}{\left(\tan x+\sqrt{3}\right)\left(\tan x-\sqrt{3}\right)}}$$
Now, multiplying the denominators as well as the numerators in respect to the given terms, we get;
$$\Rightarrow {\displaystyle \frac{1}{\tan x}}+{\displaystyle \frac{\tan x-\sqrt{3}{\tan }^{2}x-\sqrt{3}+3\tan x+\tan x+\sqrt{3}{\tan }^{2}x+\sqrt{3}+3\tan x}{{\tan }^{2}x-3}}$$
Simplifying the terms using simple algebraic expressions and methods, we get;
$$\Rightarrow {\displaystyle \frac{1}{\tan x}}+{\displaystyle \frac{2\tan x+6\tan x}{{\tan }^{2}x-3}}$$
Adding the numerator with like terms, we get;
$$\Rightarrow {\displaystyle \frac{1}{\tan x}}+{\displaystyle \frac{8\tan x}{{\tan }^{2}x-3}}$$
Now taking the L.C.M. of the denominators and multiplying the numerators with the opposite denominator, we get;
$$\Rightarrow {\displaystyle \frac{{\tan }^{2}x-3+8{\tan }^{2}x}{{\tan }^{3}x-3\tan x}}$$
Simplifying the numerator, we get;
$$\Rightarrow {\displaystyle \frac{9{\tan }^{2}x-3}{{\tan }^{3}x-3\tan x}}$$
Taking the negative sign out from both the numerator and the denominator, we get;
$$\Rightarrow {\displaystyle \frac{3-9{\tan }^{2}x}{3\tan x-{\tan }^{3}x}}$$
Taking the common terms in the numerator and denominator out, we get;
$$\Rightarrow {\displaystyle \frac{-3\left(1-3{\tan }^{2}x\right)}{-\left(3\tan x-{\tan }^{2}x\right)}}$$
Cancelling out the negative sign, we get;
$$\Rightarrow {\displaystyle \frac{3\left(1-3{\tan }^{2}x\right)}{\left(3\tan x-{\tan }^{2}x\right)}}$$
Bringing the numerator trigonometric part to the denominator for better access, we get;
$$\Rightarrow {\displaystyle \frac{3}{\left({\displaystyle \frac{3\tan x-{\tan }^{2}x}{1-3{\tan }^{2}x}}\right)}}$$
Now, the denominator is in the form of $$\tan 3x$$
$$\tan 3x={\displaystyle \frac{3\tan x-{\tan }^{2}x}{1-3{\tan }^{2}x}}$$
Substituting the denominator with the above expression, we get;
$$\Rightarrow {\displaystyle \frac{3}{\tan 3x}}$$
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,
$$\Rightarrow {\displaystyle \frac{3-9{\tan }^{2}x}{3\tan x-{\tan }^{3}x}}$$
Therefore, for the given expression, we have;
$$\cot x+\cot \left({60}^{\circ }+x\right)+\cot \left({120}^{\circ }+x\right)={\displaystyle \frac{3-9{\tan }^{2}x}{3\tan x-{\tan }^{3}x}}$$

$\therefore$ 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.