6.

Solution

Review all six trigonometric ratios: sine, cosine, tangent, cotangent, secant, & cosecant

Given,

$cot θ = 1/√3$

Using $cosec² θ − cot² θ = 1$,we can find cosec θ

$cosec θ = √(1 + cot² θ)$

$= √(1 + (1/√3)²)$

$= √(1 + (1/3)) = √((3 + 1)/3)$

$= √(4/3)$

⇒ $cosec θ = 2/√3$

So, $sin θ = 1/ cosec θ = 1/ (2/√3)$

⇒ $sin θ = √3/2$

And, we know that

$cos θ = √(1 – sin² θ)$

$= √(1 – (√3/2)²)$

$= √(1 – (3/4))$

$= √((4 – 3)/4)$

$= √(1/4)$

⇒ $cos θ = ½$

Now, using cos θ and sin θ in the expression, we have

$= 3/5$