Question

Find the value of the sum of the trigonometric functions $ \cot x+\cot \left( {{60}^{o}}+x \right)+\cot \left( {{120}^{o}}+x \right) $ ?
(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: We start solving problem by using the result \[\cot \left( A+B \right)=\dfrac{\cot \left( A \right).\cot \left( B \right)-1}{\cot \left( A \right)+\cot \left( B \right)}\] for the terms given in $ \cot x+\cot \left( {{60}^{o}}+x \right)+\cot \left( {{120}^{o}}+x \right) $ . We substitute the values required in the equations obtained and we get the answer in cotangent functions. Since the given options are present in tangent functions, we use the fact that the tangent function is inverse of the cotangent function to get the final answer.

Complete step-by-step answer:
Given that we need to find the value of $ \cot x+\cot \left( {{60}^{o}}+x \right)+\cot \left( {{120}^{o}}+x \right) $ . Let us assume the value is ‘y’.
We have got $ y=\cot x+\cot \left( {{60}^{o}}+x \right)+\cot \left( {{120}^{o}}+x \right) $ ---(1).
We know that \[\cot \left( A+B \right)=\dfrac{\cot \left( A \right).\cot \left( B \right)-1}{\cot \left( A \right)+\cot \left( B \right)}\]. Let us substitute this in equation (1).
So, we have got $ y=\cot x+\dfrac{\cot \left( {{60}^{o}} \right).\cot x-1}{\cot \left( {{60}^{o}} \right)+\cot x}+\dfrac{\cot \left( {{120}^{o}} \right).\cot x-1}{\cot \left( {{120}^{o}} \right)+\cot x} $ ---(2).
We know that the values of $ \cot \left( {{60}^{o}} \right) $ and $ \cot \left( {{120}^{o}} \right) $ are $ \dfrac{1}{\sqrt{3}} $ and $ \dfrac{-1}{\sqrt{3}} $ . Let us substitute these in equation (2).
$\Rightarrow$ $ y=\cot x+\dfrac{\dfrac{1}{\sqrt{3}}.\cot x-1}{\dfrac{1}{\sqrt{3}}+\cot x}+\dfrac{\dfrac{-1}{\sqrt{3}}.\cot x-1}{\dfrac{-1}{\sqrt{3}}+\cot x} $ .
$\Rightarrow$ $ y=\cot x+\dfrac{\dfrac{\cot x-\sqrt{3}}{\sqrt{3}}}{\dfrac{1+\sqrt{3}\cot x}{\sqrt{3}}}+\dfrac{\dfrac{-\cot x-\sqrt{3}}{\sqrt{3}}}{\dfrac{-1+\sqrt{3}\cot x}{\sqrt{3}}} $ .
$\Rightarrow$ $ y=\cot x+\dfrac{\cot x-\sqrt{3}}{1+\sqrt{3}\cot x}+\dfrac{-\cot x-\sqrt{3}}{-1+\sqrt{3}\cot x} $ .
$\Rightarrow$ \[y=\cot x+\dfrac{\left( \left( \cot x-\sqrt{3} \right)\times \left( -1+\sqrt{3}\cot x \right) \right)+\left( \left( -\cot x-\sqrt{3} \right)\times \left( 1+\sqrt{3}\cot x \right) \right)}{\left( 1+\sqrt{3}\cot x \right)\times \left( -1+\sqrt{3}\cot x \right)}\].
$\Rightarrow$ \[y=\cot x+\dfrac{\left( -\cot x+\sqrt{3}{{\cot }^{2}}x+\sqrt{3}-3\cot x \right)+\left( -\cot x-\sqrt{3}-\sqrt{3}{{\cot }^{2}}x-3\cot x \right)}{-1+\sqrt{3}\cot x-\sqrt{3}\cot x+3{{\cot }^{2}}x}\].
$\Rightarrow$ \[y=\cot x+\dfrac{-8\cot x}{-1+3{{\cot }^{2}}x}\].
$\Rightarrow$ \[y=\dfrac{\left( \cot x\times \left( -1+3{{\cot }^{2}}x \right) \right)+\left( -8\cot x \right)}{-1+3{{\cot }^{2}}x}\].
$\Rightarrow$ \[y=\dfrac{-\cot x+3{{\cot }^{3}}x-8\cot x}{-1+3{{\cot }^{2}}x}\].
$\Rightarrow$ \[y=\dfrac{3{{\cot }^{3}}x-9\cot x}{-1+3{{\cot }^{2}}x}\].
$\Rightarrow$ \[y=\dfrac{3{{\cot }^{3}}x-9\cot x}{3{{\cot }^{2}}x-1}\]---(3).
We know that $ \cot x=\dfrac{1}{\tan x} $ . Let us substitute this in equation (3).
$\Rightarrow$ \[y=\dfrac{\left( \dfrac{3}{{{\tan }^{3}}x} \right)-\left( \dfrac{9}{\tan x} \right)}{-1+\left( \dfrac{3}{{{\tan }^{2}}x} \right)}\].
$\Rightarrow$ \[y=\dfrac{\left( \dfrac{3-9{{\tan }^{2}}x}{{{\tan }^{3}}x} \right)}{\left( \dfrac{-{{\tan }^{2}}x+3}{{{\tan }^{2}}x} \right)}\].
$\Rightarrow$ \[y=\dfrac{\left( \dfrac{3-9{{\tan }^{2}}x}{\tan x} \right)}{-{{\tan }^{2}}x+3}\].
$\Rightarrow$ \[y=\dfrac{3-9{{\tan }^{2}}x}{\left( \tan x \right)\times \left( -{{\tan }^{2}}x+3 \right)}\].
$\Rightarrow$ \[y=\dfrac{3-9{{\tan }^{2}}x}{3\tan x-{{\tan }^{3}}x}\].
∴ The value of $ \cot x+\cot \left( {{60}^{o}}+x \right)+\cot \left( {{120}^{o}}+x \right) $ is \[\dfrac{3-9{{\tan }^{2}}x}{3\tan x-{{\tan }^{3}}x}\].
So, the correct answer is “Option D”.

Note: Alternatively, we can check the options by assigning a value for ‘x’ which is between $ {{0}^{o}} $ and $ {{90}^{o}} $. Similarly, we can expect tangent, cosine and sine functions instead of given cotangent functions. Whatever the trigonometric function provided in the problem, we start and solve the problem by using the same procedure.