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Question: If \[\sum _{m = 1}^n{\tan ^{ - 1}}\left[ {\dfrac{{2m}}{{{m^4} + {m^2} + 2}}} \right]\] is equal to ...

If ∑m=1ntan⁡−1[2mm4+m2+2]\sum _{m = 1}^n{\tan ^{ - 1}}\left[ {\dfrac{{2m}}{{{m^4} + {m^2} + 2}}} \right] is equal to
A.tan⁡−1[2mn2+n+2]{\tan ^{ - 1}}\left[ {\dfrac{{2m}}{{{n^2} + n + 2}}} \right]
B.tan⁡−1[n2+nn2+n+2]{\tan ^{ - 1}}\left[ {\dfrac{{{n^2} + n}}{{{n^2} + n + 2}}} \right]
C.tan⁡−1[n2+n+2n2+n]{\tan ^{ - 1}}\left[ {\dfrac{{{n^2} + n + 2}}{{{n^2} + n}}} \right]
D.None of these

Explanation

Solution

Hint : Use the trigonometric identity of tan⁡−1x±tan⁡−1y{\tan ^{ - 1}}x \pm {\tan ^{ - 1}}y as tan⁡−1y(x±y1±xy){\tan ^{ - 1}}y\left( {\dfrac{{x \pm y}}{{1 \pm xy}}} \right) and compute the summation as given in the question. Use the method of factorization in order to change the numerator of the given expression. After simplifying the summation we will be left with a few terms then use the suitable trigonometric formula to simplify the answer.

Complete step-by-step answer :
We are to find the value of ∑m=1ntan⁡−1[2mm4+m2+2]\sum _{m = 1}^n{\tan ^{ - 1}}\left[ {\dfrac{{2m}}{{{m^4} + {m^2} + 2}}} \right] as asked in the question.
In general the formula for tan⁡−1x±tan⁡−1y{\tan ^{ - 1}}x \pm {\tan ^{ - 1}}y is tan⁡−1y(x±y1±xy){\tan ^{ - 1}}y\left( {\dfrac{{x \pm y}}{{1 \pm xy}}} \right) … (1)
Now given summation can be rewritten as ∑m=1ntan⁡−1[2mm4+m2+2]=∑m=1ntan⁡−1[2m1+(m4+m2+1)]\sum\limits_{m = 1}^n {{{\tan }^{ - 1}}\left[ {\dfrac{{2m}}{{{m^4} + {m^2} + 2}}} \right]} = \sum\limits_{m = 1}^n {{{\tan }^{ - 1}}\left[ {\dfrac{{2m}}{{1 + \left( {{m^4} + {m^2} + 1} \right)}}} \right]}
Now, on factorization of the denominator we get m4+m2+1=(m2+m+1)(m2−m+1){m^4} + {m^2} + 1 = \left( {{m^2} + m + 1} \right)\left( {{m^2} - m + 1} \right) and difference between (m2+m+1)\left( {{m^2} + m + 1} \right) and (m2−m+1)\left( {{m^2} - m + 1} \right) is 2m2m.
Therefore the expression becomes ∑m=1ntan⁡−1[2m1+(m4+m2+1)]=∑m=1ntan⁡−1[(m2+m+1)−(m2−m+1)1+(m2+m+1)(m2−m+1)]                     ...(2)\sum\limits_{m = 1}^n {{{\tan }^{ - 1}}\left[ {\dfrac{{2m}}{{1 + \left( {{m^4} + {m^2} + 1} \right)}}} \right]} = \sum\limits_{m = 1}^n {{{\tan }^{ - 1}}\left[ {\dfrac{{\left( {{m^2} + m + 1} \right) - \left( {{m^2} - m + 1} \right)}}{{1 + \left( {{m^2} + m + 1} \right)\left( {{m^2} - m + 1} \right)}}} \right]} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)
Comparing (1) and (2) we get
∑m=1ntan⁡−1[2mm4+m2+2]=∑m=1n[tan⁡−1(m2+m+1)−tan⁡−1(m2−m+1)]\sum\limits_{m = 1}^n {{{\tan }^{ - 1}}\left[ {\dfrac{{2m}}{{{m^4} + {m^2} + 2}}} \right]} = \sum\limits_{m = 1}^n {\left[ {{{\tan }^{ - 1}}\left( {{m^2} + m + 1} \right) - {{\tan }^{ - 1}}\left( {{m^2} - m + 1} \right)} \right]}
Now putting value from 1,2,...,n1,2,...,n to open the summation we get
=tan⁡−1(m2+1+1)−tan⁡−1(m2−1+1)+tan⁡−1(22+2+1)−tan⁡−1(22−2+1)+...+tan⁡−1(n2+n+1)−tan⁡−1(n2−n+1)= {\tan ^{ - 1}}\left( {{m^2} + 1 + 1} \right) - {\tan ^{ - 1}}\left( {{m^2} - 1 + 1} \right) + {\tan ^{ - 1}}\left( {{2^2} + 2 + 1} \right) - {\tan ^{ - 1}}\left( {{2^2} - 2 + 1} \right) + ... + {\tan ^{ - 1}}\left( {{n^2} + n + 1} \right) - {\tan ^{ - 1}}\left( {{n^2} - n + 1} \right)
Simplify the expression to make the terms simple and easy to calculate we get ,
=tan⁡−1(3)−tan⁡−1(1)+tan⁡−1(7)−tan⁡−1(3)...+tan⁡−1(n2+n+1)−tan⁡−1(n2−n+1)= {\tan ^{ - 1}}(3) - {\tan ^{ - 1}}(1) + {\tan ^{ - 1}}(7) - {\tan ^{ - 1}}(3)... + {\tan ^{ - 1}}\left( {{n^2} + n + 1} \right) - {\tan ^{ - 1}}\left( {{n^2} - n + 1} \right)
Here alternative terms will cancel out and we are left with tan⁡−1(n2+n+1)−tan⁡−1(1){\tan ^{ - 1}}\left( {{n^2} + n + 1} \right) - {\tan ^{ - 1}}(1)
Therefore after simplification we get ∑m=1ntan⁡−1[2mm4+m2+2]=tan⁡−1(n2+n+1)−tan⁡−1(1)\sum\limits_{m = 1}^n {{{\tan }^{ - 1}}\left[ {\dfrac{{2m}}{{{m^4} + {m^2} + 2}}} \right]} = {\tan ^{ - 1}}\left( {{n^2} + n + 1} \right) - {\tan ^{ - 1}}(1)
= {\tan ^{ - 1}}\left[ {\dfrac{{\left( {{n^2} + n + 1} \right) - \left( 1 \right)}}{{1 + \left( {{n^2} + n + 1} \right)\left( 1 \right)}}} \right]$$$$ = {\tan ^{ - 1}}\left[ {\dfrac{{{n^2} + n}}{{{n^2} + n + 2}}} \right]
Hence we get our answer to the question.

Note : The angles of the sine, the cosine, and the tangent are the primary classification of functions of trigonometry. And the three functions which are the cotangent, the secant and the cosecant can be derived from the primary functions. You need to have the grip of various trigonometric formulas in order to attempt such questions. In mathematics, the inverse trigonometric functions are used to obtain an angle from any of the angle's trigonometric ratios.