(A) 1/√3 (B) √3 (C) 1 (D) 0
(C) 1
As indicated by the inquiry,
for example 9α is an intense point
We realize that,
In this way,
Since,
Subsequently, sin (90°-9∝) = sin∝
10∝ = 90°
∝ = 9°
Subbing ∝ = 9° in tan 5∝, we get,
∴, tan 5∝ = 1