The value of sec−1(41∑k=010sec(127π+2kπ)sec(127π+2(k+1)π)) in the interval [−4π,43π] equals
Answer: 0
Solution
Using the identity secAsecB=sin(B−A)1(tanB−tanA) with B−A=2π, the sum telescopes: ∑k=010sec(127π+2kπ)sec(127π+2(k+1)π)=tan(211π+127π)−tan127π=−cot127π−tan127π=−sin67π2=4. Thus sec−1(1)=0.00.
JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Numerical answer