The correct option (c) (π/4)
Explanation:
∵ sin–1(x) = cosec–1(1/x) = tan–1[x /√(1 – x2)]
∴ cosec–1√5 = sin–1(1/√5) = tan–1(1/2)
∵ cosec–1√5 + cosec–1√(65) + cosec–1√(325) + ....... ∞
= limn➙∞[tan–1(1/2) + tan–1(1/8) + tan–1(1/18) + .….. nth term]
= limn➙∞[tan–1{(3 – 1)/(1 + 3)} + tan–1{(5 – 3)/(1 + 15)} + tan–1{(7 – 5)/(1 + 35)} + …… tan–1{[(2n + 1) – (2n – 1)]/[1 + (2n – 1)(2n + 1)]}]
= limn➙∞[tan–1(2n + 1) – tan–11]
= (π/2) – (π/4)
= (π/4).
