AP EAMCET201921 Apr 2019Evening ShiftMathematicsInverse Trigonometric FunctionsActual
If ∑ k = 1 n tan - 1 1 k 2 + k + 1 = tan - 1 θ , then θ =
Options
- An n + 2
- Bn n + 1
- C1
- Dn n - 1
Correct answer
A. n n + 2
Step-by-step solution
It is given that, ∑ k = 1 n tan - 1 1 k 2 + k + 1 = tan - 1 θ Simplifying the above expression we get, ∑ k = 1 n tan - 1 ( k + 1 ) - k 1 + k ( k + 1 ) = tan - 1 θ ⇒ ∑ k = 1 n tan - 1 ( k + 1 ) - tan - 1 k = tan - 1 θ tan - 1 2 - tan - 1 1 + tan - 1 3 - tan - 1 2 + tan - 1 4 - tan - 1 3 + … + tan - 1 ( n + 1 ) - tan - 1 n = tan - 1 θ ⇒ tan - 1 ( n + 1 ) - tan - 1 1 = tan - 1 θ ⇒ tan - 1 n + 1 - 1 1 + n + 1 = tan - 1 θ ⇒ n 2 + n = θ