(A) Let f₁(x) = cosx

In interval,

Therefore, f₁(x) = cosx is strictly decreasing in interval.

(B) Let f₂(x) = cos2x

Now, 0 < x <

⇒ 0 < 2x < π

⇒ sin2x > 0

⇒ -2sin2x < 0

Therefore, f₂(x) = cos2x is strictly decreasing in interval.

(C) Let f₃(x) = cos3x

Now,

⇒ sin3x = 0

⇒ 3x = π, as xϵ

⇒ x =

The point x = divides the interval into two distinct intervals.

i.e. and

Now, in interval, ,

f₃'(x) = -3sin3x < 0 as (0 < x < => 0 < 3x < π)

Therefore, f₃ is strictly decreasing in interval.

Now, in interval

f₃'(x)=-3sin3x > 0 as

Therefore, f₃ is strictly increasing in interval.

(D) Let f₄ = tanx

In interval,

Therefore, f4 is strictly increasing in interval .