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Question: The value of \(\cot \sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k}...

The value of cot⁡∑n=123cot⁡−1(1+∑k=1n2k)\cot \sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} is
(a)2325\left( a \right)\dfrac{{23}}{{25}}
(b)2523\left( b \right)\dfrac{{25}}{{23}}
(c)2324\left( c \right)\dfrac{{23}}{{24}}
(d)2423\left( d \right)\dfrac{{24}}{{23}}

Explanation

Solution

In this particular question use the concept that cot⁡−1x=tan⁡−1(1x){\cot ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right) and use the concept that the summation of first n natural number is ∑k=1nk=n(n+1)2\sum\limits_{k = 1}^n k = \dfrac{{n\left( {n + 1} \right)}}{2} so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given equation:
cot⁡∑n=123cot⁡−1(1+∑k=1n2k)\cot \sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} .......................... (1)
Now first find the value of
∑n=123cot⁡−1(1+∑k=1n2k)\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right)}
Now above equation is also written as
⇒∑n=123cot⁡−1(1+2∑k=1nk)\Rightarrow \sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + 2\sum\limits_{k = 1}^n k } \right)}
Now as we know that the summation of first n natural number is ∑k=1nk=n(n+1)2\sum\limits_{k = 1}^n k = \dfrac{{n\left( {n + 1} \right)}}{2} so we have,
⇒∑n=123cot⁡−1(1+2n(n+1)2)\Rightarrow \sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + 2\dfrac{{n\left( {n + 1} \right)}}{2}} \right)}
⇒∑n=123cot⁡−1(1+n(n+1))\Rightarrow \sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + n\left( {n + 1} \right)} \right)}
Now as we know that cot⁡−1x=tan⁡−1(1x){\cot ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right) so use this property in the above equation we have,
⇒∑n=123tan⁡−1(11+n(n+1))\Rightarrow \sum\limits_{n = 1}^{23} {{{\tan }^{ - 1}}\left( {\dfrac{1}{{1 + n\left( {n + 1} \right)}}} \right)}
Now the above equation is also written as
⇒∑n=123tan⁡−1((n+1)−11+n(n+1))\Rightarrow \sum\limits_{n = 1}^{23} {{{\tan }^{ - 1}}\left( {\dfrac{{\left( {n + 1} \right) - 1}}{{1 + n\left( {n + 1} \right)}}} \right)}
Now as we know that tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy){\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right), so use this property in the above equation we have, where x = n + 1, y = n
⇒∑n=123tan⁡−1((n+1)−11+n(n+1))=∑n=123(tan⁡−1(n+1)−tan⁡−1n)\Rightarrow \sum\limits_{n = 1}^{23} {{{\tan }^{ - 1}}\left( {\dfrac{{\left( {n + 1} \right) - 1}}{{1 + n\left( {n + 1} \right)}}} \right)} = \sum\limits_{n = 1}^{23} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)}
Now expand the summation we have,
⇒∑n=123(tan⁡−1(n+1)−tan⁡−1n)=(tan⁡−12−tan⁡−11)+(tan⁡−13−tan⁡−12)+(tan⁡−14−tan⁡−13)+...... \+(tan⁡−123−tan⁡−122)+(tan⁡−124−tan⁡−123) \begin{gathered} \Rightarrow \sum\limits_{n = 1}^{23} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)} = \left( {{{\tan }^{ - 1}}2 - {{\tan }^{ - 1}}1} \right) + \left( {{{\tan }^{ - 1}}3 - {{\tan }^{ - 1}}2} \right) + \left( {{{\tan }^{ - 1}}4 - {{\tan }^{ - 1}}3} \right) + ...... \\\ \+ \left( {{{\tan }^{ - 1}}23 - {{\tan }^{ - 1}}22} \right) + \left( {{{\tan }^{ - 1}}24 - {{\tan }^{ - 1}}23} \right) \\\ \end{gathered}
So as we see that all the terms are cancel out except two terms so we have,
⇒∑n=123(tan⁡−1(n+1)−tan⁡−1n)=tan⁡−124−tan⁡−11\Rightarrow \sum\limits_{n = 1}^{23} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)} = {\tan ^{ - 1}}24 - {\tan ^{ - 1}}1
Now again use that tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy){\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) so we have,
⇒∑n=123cot⁡−1(1+∑k=1n2k)=tan⁡−1(24−11+24(1))=tan⁡−1(2325)=cot⁡−1(2523)\Rightarrow \sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} = {\tan ^{ - 1}}\left( {\dfrac{{24 - 1}}{{1 + 24\left( 1 \right)}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{23}}{{25}}} \right) = {\cot ^{ - 1}}\left( {\dfrac{{25}}{{23}}} \right)
Now substitute this value in equation (1) we have,
⇒cot⁡∑n=123cot⁡−1(1+∑k=1n2k)=cot⁡(cot⁡−12523)=2523\Rightarrow \cot \sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} = \cot \left( {{{\cot }^{ - 1}}\dfrac{{25}}{{23}}} \right) = \dfrac{{25}}{{23}}
So this is the required answer.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic trigonometric identity such as tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy){\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) and rest of the all are stated above, so first simplify the given equation as above then apply this property and simplify as above and check which of the terms are cancel out as above then again apply this property we will get the required answer.