Question

Simplify the expressions of the sums ${{\cot }^{2}}\dfrac{\pi }{2n+1}+{{\cot }^{2}}\dfrac{2\pi }{2n+1}+{{\cot }^{2}}\dfrac{3\pi }{2n+1}+......+{{\cot }^{2}}\dfrac{n\pi }{2n+1}=$
A. $\dfrac{n\left( 2n+1 \right)}{3}$
B. $\dfrac{n\left( 2n+1 \right)}{6}$
C. $\dfrac{n\left( 2n-1 \right)}{3}$
D. $\dfrac{n\left( 2n-1 \right)}{6}$

Answer 1

Hint: For solving this question you should know about the sum of trigonometric functions. In this problem we will use the formula of $\sin 2a$ expanding formula and then by putting this as zero, we will get the value for the angle and using these we will expand them in the form of ${{\cot }^{2}}$ terms sums and then will find the sum of the roots of the equations.

Complete step by step answer:
According to our question it is asked to us to solve the expression of sums. Our expression is given as: ${{\cot }^{2}}\dfrac{\pi }{2n+1}+{{\cot }^{2}}\dfrac{2\pi }{2n+1}+{{\cot }^{2}}\dfrac{3\pi }{2n+1}+......+{{\cot }^{2}}\dfrac{n\pi }{2n+1}$.
As we know that any complex number like ${{e}^{ix}}=\cos x+i\sin x$ can be expressed as ${{e}^{inx}}={{\left( \cos x+i\sin x \right)}^{n}}=\cos nx+i\sin nx$
Now, we know that for expansion of any term having some power we use binomial expansion. So, we will use binomial expansion that is ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$ to expand ${{\left( \cos x+i\sin x \right)}^{n}}$. Therefore, we get
\[{{\left( \cos x+i\sin x \right)}^{n}}={}^{n}{{C}_{0}}{{\left( \cos x \right)}^{n}}{{\left( i\sin x \right)}^{0}}+{}^{n}{{C}_{1}}{{\left( \cos x \right)}^{n-1}}{{\left( i\sin x \right)}^{1}}+...+{}^{n}{{C}_{n}}{{\left( \cos x \right)}^{0}}{{\left( i\sin x \right)}^{n}}\]
Therefore, we can say
\[{}^{n}{{C}_{0}}{{\left( \cos x \right)}^{n}}{{\left( i\sin x \right)}^{0}}+{}^{n}{{C}_{1}}{{\left( \cos x \right)}^{n-1}}{{\left( i\sin x \right)}^{1}}+...+{}^{n}{{C}_{n}}{{\left( \cos x \right)}^{0}}{{\left( i\sin x \right)}^{n}}=\cos nx+i\sin nx\]
And on comparing imaginary terms of both side, we will get
\[\begin{align}
  & \sin \left( n \right)x={}^{n}{{C}_{1}}{{\cos }^{n-1}}x\sin x{{-}^{n}}{{C}_{3}}{{\cos }^{n-3}}x{{\sin }^{3}}x{{+}^{n}}{{C}_{5}}{{\cos }^{n-5}}x{{\sin }^{5}}x-....= \\
 & \text{ }{{\sin }^{n}}x\left[ ^{n}{{C}_{1}}{{\cot }^{n}}x{{-}^{n}}{{C}_{3}}{{\cot }^{n-3}}x{{+}^{n}}{{C}_{5}}{{\cot }^{n-5}}x-... \right] \\
\end{align}\]
Now, if we replace n by 2n+1 in above equation and x by $\alpha $, we will get the equation as
$\begin{align}
  & \sin \left( 2n+1 \right)\alpha ={}^{2n+1}{{C}_{1}}{{\cos }^{2n}}\alpha \sin {{\alpha }^{-2n+1}}{{C}_{3}}{{\cos }^{2n-2}}\alpha {{\sin }^{3}}\alpha {{+}^{2n+1}}{{C}_{5}}{{\cos }^{2n-4}}\alpha {{\sin }^{5}}\alpha -....= \\
 & \text{ }{{\sin }^{2n+1}}\alpha \left[ ^{2n+1}{{C}_{1}}{{\cot }^{2n}}\alpha {{-}^{2n+1}}{{C}_{3}}{{\cot }^{2n-2}}\alpha {{+}^{2n+1}}{{C}_{5}}{{\cot }^{2n-4}}\alpha -... \right]\ldots \ldots \ldots \left( i \right) \\
\end{align}$
So, we will use this for solving our problem, by writing equation (i) we can write,
$\operatorname{Sin}\left( 2n+1 \right)\alpha ={{\sin }^{2n+1}}\alpha \left[ ^{2n+1}{{C}_{1}}{{\cot }^{2n}}\alpha {{-}^{2n+1}}{{C}_{3}}{{\cot }^{2n-2}}\alpha {{+}^{2n+1}}{{C}_{5}}{{\cot }^{2n-4}}\alpha -... \right]\ldots $
Now if $LHS=\operatorname{Sin}\left( 2n+1 \right)\alpha =0$, when $\left( 2n+1 \right)=\pi \cdot r\text{ where }r\in N$ or $\alpha =\dfrac{\pi }{2n+1},\dfrac{2\pi }{2n+1},\dfrac{3\pi }{2n+1},......$
Hence, the equation (i) gives us,
$^{2n+1}{{C}_{1}}{{\cot }^{2n}}\alpha {{-}^{2n+1}}{{C}_{3}}{{\cot }^{2n-2}}\alpha {{+}^{2n+1}}{{C}_{5}}{{\cot }^{2n-4}}\alpha -...=0$
The above is an ${{n}^{th}}$ degree equation in ${{\cot }^{2}}\alpha $,
Hence,
${{\cot }^{2}}\dfrac{\pi }{2n+1},{{\cot }^{2}}\dfrac{2\pi }{2n+1},{{\cot }^{2}}\dfrac{3\pi }{2n+1},{{\cot }^{2}}\dfrac{n\pi }{2n+1}$ are the $n$ roots of the equation.
$^{2n+1}{{C}_{1}}{{x}^{n}}{{-}^{2n+1}}{{C}_{3}}{{x}^{n-1}}{{+}^{2n+1}}{{C}_{5}}{{x}^{n-2}}-...=0\ldots \ldots \ldots \left( ii \right)$
So, sum of the roots of equation (ii),
$\begin{align}
  & =-\dfrac{\text{coefficient of }{{x}^{n-1}}}{\text{coefficient of }{{x}^{n}}}=\dfrac{{}^{2n+1}{{C}_{3}}}{{}^{2n+1}{{C}_{1}}} \\
 & =\dfrac{\left( 2n+1 \right).2n\left( 2n-1 \right)}{\left( 1.2.3 \right)\left( 2n+1 \right)}=\dfrac{n\left( 2n-1 \right)}{3} \\
\end{align}$

So, the correct answer is “Option C”.

Note: While solving these type of equations you have to use the other forms of the trigonometric functions for solving to any other function and you have to be careful for finding the roots for their functions, because these depend on the trigonometric functions and the solution depends on these.