Question

Calculate \[\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} } \right)\]

Answer 1

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

Complete step-by-step answer:
Given equation is \[\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} } \right)\]
In order to find this expression, first we have to derive
\[\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} \]
Now, we will find expression \[\sum\limits_{p = 1}^n {2p} \]
We know that the range of p is from 1 to n.
\[\sum\limits_{p = 1}^n {2p} = 2(1 + 2 + 3 + .. + n)\]
We can rewrite this equation as,
\[ = 2\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)\]
After solving this equation, we get
\[ = ({n^2} + n)\]
Now, we can derive
\[\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} \]
\[ = \sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + ({n^2} + n)} \right)} \]
\[ = \sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)} \]
But, we know \[{\cot ^{ - 1}}A = {\tan ^{ - 1}}\dfrac{1}{A}\]
\[ = \sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right)} \]
We can rewrite denominator of this equation as,
\[ = \sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {\dfrac{1}{{1 + n(n + 1)}}} \right)} \]
Now, we can rewrite numerator as follows,
\[ = \sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {\dfrac{{(n + 1) - n}}{{1 + n(n + 1)}}} \right)} \]
But we know,
\[{\tan ^{ - 1}}\left( x \right) - {\tan ^{ - 1}}\left( y \right) = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)\]
Here, x = (n+1) and y = n
\[ = \sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {n + 1} \right)} - {\tan ^{ - 1}}\left( n \right)\]
As range is given from n = 1 to 19, we can write this expression as,
\[\begin{array}{l}
 = \left( {{{\tan }^{ - 1}}\left( 2 \right) - {{\tan }^{ - 1}}\left( 1 \right)} \right) + \left( {{{\tan }^{ - 1}}\left( 2 \right) - {{\tan }^{ - 1}}\left( 1 \right)} \right)\\
 + ... + \left( {{{\tan }^{ - 1}}\left( {20} \right) - {{\tan }^{ - 1}}\left( {19} \right)} \right)
\end{array}\]
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.