(A) We have
tan−1() =
tan−1() =
tan−1() =
tan−1(| (2i+1)−(2i−1) |
| 1+(2i−1)(2i+1) |
) =
(tan−1(2i+1)−tan−1(2i+1) =
+ L +
tan−1(2n+1)−tan−1(2n−1) =
(tan−1(2n+1)−tan−1) =
− =
Therefore, (A)→(Q).
(B) We have
cos
θ1 =
=
=
On applying componendo and dividendo, we get
=
tan2() =
Similarly,
tan2 =
tan2+tan2 =
+
Now, we see that a, b, c are in A.P. Therefore,
2b = a + c
tan2+tan2 =
=
=
Therefore, (B)→(P).
(C) Equation of the line is
=
=
and the line through the origin is
=
=
= λ
which is perpendicular to given line a + 2b + 2c = 0. Now, P(aλ, bλ, cλ) lies on given line
aλ =
=
⇒ a =
⇒ c = 2a
Therefore,
bλ - 1 = cλ ⇒ λ =
() That is,
a + 2b + 4a = 0 ⇒ b =
− 2aλ = bλ - 1
λ =
() λ =
=
=
Therefore,
P (aλ , bλ , cλ) =
=
P(,−,−) Therefore, the perpendicular distance is
√++ =
√ =
√ =
Therefore, (C)→(R).