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Question: Calculate \[\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n...

Calculate cot⁡(∑n=119cot⁡−1(1+∑p=1n2p))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} } \right)

Explanation

Solution

This question is from Inverse Trigonometric Functions. Firstly, we have to solve ∑p=1n2p\sum\limits_{p = 1}^n {2p} . Then, we can calculate ∑n=119cot⁡−1(1+∑p=1n2p)\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} by using inverse trigonometric properties as follows.
tan⁡−1A=cot⁡−1(1A){\tan ^{ - 1}}A = {\cot ^{ - 1}}\left( {\dfrac{1}{A}} \right)
tan⁡−1A−tan⁡−1B=tan⁡−1(A−B1+AB);AB>−1{\tan ^{ - 1}}A - {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A - B}}{{1 + AB}}} \right);AB > - 1.

Complete step-by-step answer:
Given equation is cot⁡(∑n=119cot⁡−1(1+∑p=1n2p))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} } \right)
In order to find this expression, first we have to derive
∑n=119cot⁡−1(1+∑p=1n2p)\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2p} } \right)}
Now, we will find expression ∑p=1n2p\sum\limits_{p = 1}^n {2p}
We know that the range of p is from 1 to n.
∑p=1n2p=2(1+2+3+..+n)\sum\limits_{p = 1}^n {2p} = 2(1 + 2 + 3 + .. + n)
We can rewrite this equation as,
=2(n(n+1)2)= 2\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)
After solving this equation, we get
=(n2+n)= ({n^2} + n)
Now, we can derive
∑n=119cot⁡−1(1+∑p=1n2p)\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2p} } \right)}
=∑n=119cot⁡−1(1+(n2+n))= \sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + ({n^2} + n)} \right)}
=∑n=119cot⁡−1(n2+n+1)= \sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)}
But, we know cot⁡−1A=tan⁡−11A{\cot ^{ - 1}}A = {\tan ^{ - 1}}\dfrac{1}{A}
=∑n=119tan⁡−1(1n2+n+1)= \sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right)}
We can rewrite denominator of this equation as,
=∑n=119tan⁡−1(11+n(n+1))= \sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {\dfrac{1}{{1 + n(n + 1)}}} \right)}
Now, we can rewrite numerator as follows,
=∑n=119tan⁡−1((n+1)−n1+n(n+1))= \sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {\dfrac{{(n + 1) - n}}{{1 + n(n + 1)}}} \right)}
But we know,
tan⁡−1(x)−tan⁡−1(y)=tan⁡−1(x−y1+xy){\tan ^{ - 1}}\left( x \right) - {\tan ^{ - 1}}\left( y \right) = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)
Here, x = (n+1) and y = n
=∑n=119tan⁡−1(n+1)−tan⁡−1(n)= \sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {n + 1} \right)} - {\tan ^{ - 1}}\left( n \right)
As range is given from n = 1 to 19, we can write this expression as,

= \left( {{{\tan }^{ - 1}}\left( 2 \right) - {{\tan }^{ - 1}}\left( 1 \right)} \right) + \left( {{{\tan }^{ - 1}}\left( 2 \right) - {{\tan }^{ - 1}}\left( 1 \right)} \right)\\\ \+ ... + \left( {{{\tan }^{ - 1}}\left( {20} \right) - {{\tan }^{ - 1}}\left( {19} \right)} \right) \end{array}$$ Now, this equation becomes, $$ = {\tan ^{ - 1}}\left( {20} \right) - {\tan ^{ - 1}}\left( 1 \right)$$ Now, according to inverse trigonometry properties $${\tan ^{ - 1}}\left( x \right) - {\tan ^{ - 1}}\left( y \right) = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)$$ Here, x = 20 and y = 1 Now, we get $$ = {\tan ^{ - 1}}\left( {\dfrac{{20 - 1}}{{1 + 20 \times 1}}} \right)$$ $$ = {\tan ^{ - 1}}\left( {\dfrac{{19}}{{21}}} \right)$$ Thus given expression reduces to $$ = \cot \left( {{{\tan }^{ - 1}}\left( {\dfrac{{19}}{{21}}} \right)} \right)$$ But , we have $${\cot ^{ - 1}}\left( {\dfrac{A}{B}} \right) = {\tan ^{ - 1}}\left( {\dfrac{B}{A}} \right)$$ Now, equation becomes $$ = \cot \left( {{{\cot }^{ - 1}}\left( {\dfrac{{21}}{{19}}} \right)} \right)$$ Now, we know that $$\cot \left( {{{\cot }^{ - 1}}\left( A \right)} \right) = A$$ Thus, equation can be reduced as, $$ = \dfrac{{21}}{{19}}$$ This is the required solution. **So, the correct answer is “Option C”.** **Note:** In this problem, students must take care of the range given. It has been seen that as soon as students see $$\dfrac{{19}}{{21}}$$, they instantly mark this answer and forget to convert $$tan$$ expression into $$cot$$ expression. Here, option (d) is incorrect. One can go wrong while using inverse trigonometric formulas. All $$cot$$ expressions must need to convert to $$tan$$ expressions. Then, students must use inverse trigonometric properties. Calculation plays an important role in these types of trigonometric problems.