Answer: Given, sin θ = √2/6
We know that sin^2θ + cos^2θ = 1
Solving for cos θ, we get:
cos θ = ±√(1 - sin^2θ)
cos θ = ±√(1 - (2/6)^2)
cos θ = ±√(1 - 1/9)
cos θ = ±√(8/9)
Since θ is acute, cos θ is positive. Hence,
cos θ = √(8/9) = (2√2)/3
We know that tan θ = sin θ / cos θ
tan θ = (√2/6) / [(2√2)/3]
tan θ = √2/4
We know that cot θ = 1/tan θ
cot θ = 1/ (√2/4) = (2√2)/2 = √2
We know that csc θ = 1/ sin θ
csc θ = 1/ (√2/6) = (6√2)/2 = 3√2
We know that sec θ = 1/ cos θ
sec θ = 1/[(2√2)/3] = (3√2)/2
Hence, the exact values of the remaining trigonometric functions of the angle θ are:
cos θ = (2√2)/3
tan θ = √2/4
cot θ = √2
csc θ = 3√2
sec θ = (3√2)/2