Question

If \[y = {\tan ^{ - 1}}\dfrac{1}{{{x^2} + x + 1}} + {\tan ^{ - 1}}\dfrac{1}{{{x^2} + 3x + 3}} + {\tan ^{ - 1}}\dfrac{1}{{{x^2} + 5x + 7}} + \] ………. to \[n\] terms, then
A) \[\dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {{\left( {x + n} \right)}^2}}} - \dfrac{1}{{1 + {x^2}}}\] ​
B) \[\dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\left( {x + n} \right)}^2}}} - \dfrac{1}{{1 + {x^2}}}\]
C) \[\dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {{\left( {x + n} \right)}^2}}} + \dfrac{1}{{1 + {x^2}}}\]
D) None of these

Answer 1

Hint:
Here we need to find the differentiation of a given trigonometric equation. For that, we will break the terms inside the bracket. Then we will use the basic inverse trigonometric formulas to simplify the terms further. We will then differentiate the simplified trigonometric equation with respect to the given variable. After differentiating each term, we will get the final answer.

Complete step by step solution:
The given trigonometric equation is
\[y = {\tan ^{ - 1}}\dfrac{1}{{{x^2} + x + 1}} + {\tan ^{ - 1}}\dfrac{1}{{{x^2} + 3x + 3}} + {\tan ^{ - 1}}\dfrac{1}{{{x^2} + 5x + 7}} + \] ………. to \[n\]

Now, we will break the term of numerator of all the terms. , the quadratic equations are used in the denominator. So, we will factorize these quadratic equations used in the denominator of all the terms.
\[ \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + x\left( {x + 1} \right)}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + {x^2} + 2x + x + 2}}} \right) + .....{\tan ^{ - 1}}\left( {\dfrac{1}{{1 + \left( {x + n - 1} \right)\left( {x + n} \right)}}} \right)\]
On further simplification, we get
\[ \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + x\left( {x + 1} \right)}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + \left( {x + 1} \right)\left( {x + 2} \right)}}} \right) + .....{\tan ^{ - 1}}\left( {\dfrac{1}{{1 + \left( {x + n - 1} \right)\left( {x + n} \right)}}} \right)\]
Now, we will break the terms inside the bracket, we get
\[ \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{\left( {x + 1} \right) - x}}{{1 + x\left( {x + 1} \right)}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{\left( {x + 2} \right) - \left( {x + 1} \right)}}{{1 + \left( {x + 1} \right)\left( {x + 2} \right)}}} \right) + .... + {\tan ^{ - 1}}\left( {\dfrac{{\left( {x + n} \right) - \left( {x + n - 1} \right)}}{{1 + \left( {x + n - 1} \right)\left( {x + n} \right)}}} \right)\]
We know the inverse trigonometric formula;
\[{\tan ^{ - 1}}\left( {\dfrac{{A - B}}{{1 + A.B}}} \right) = {\tan ^{ - 1}}A - {\tan ^{ - 1}}B\]
Using this formula for each term, we get
\[ \Rightarrow y = {\tan ^{ - 1}}\left( {x + 1} \right) - {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {x + 2} \right) - {\tan ^{ - 1}}\left( {x + 1} \right) + .... + {\tan ^{ - 1}}\left( {x + n} \right) - {\tan ^{ - 1}}\left( {x + n - 1} \right)\]
On subtracting the same terms, we get
\[ \Rightarrow y = {\tan ^{ - 1}}\left( {x + n} \right) - {\tan ^{ - 1}}x\]
Now, we will differentiate both sides of equation with respect to \[x\]
\[\begin{array}{l} \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {{{\tan }^{ - 1}}\left( {x + n} \right) - {{\tan }^{ - 1}}x} \right)}}{{dx}}\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d{{\tan }^{ - 1}}\left( {x + n} \right)}}{{dx}} - \dfrac{{d{{\tan }^{ - 1}}x}}{{dx}}\end{array}\]
On differentiating each term, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {{\left( {x + n} \right)}^2}}} - \dfrac{1}{{1 + {x^2}}}\]

Hence, the correct option is option A.

Note:
Here we have used the basic trigonometric formulas to simplify the trigonometric equation. The inverse trigonometric functions are also called as the anti trigonometric functions or sometimes they are also called as arcus functions or cyclometric functions. The inverse trigonometric functions of cosine, sine, cosecant , tangent, secant and cotangent are used to find the angle of a triangle. It is used in many fields, for example - engineering, geometry, physics, etc. Inverse trigonometric formulas make the solution easy and short.