If $cosec^2θ + cot^2 θ = \frac{5}{3}$, then what is the value of cot2θ ?

$-\frac{1}{\sqrt{3}}$

cosec²θ + cot²θ = \(\frac{5}{3}\)     ----(1)

cosec²θ - cot²θ = 1    ----(2)

On subtracting equation 2 from equation 1.

2 cot²θ  = \(\frac{5}{3}\) - 1 

cot²θ  = \(\frac{1}{3}\)

cotθ  = \(\frac{1}{√3}\)

{ we know, cot60º  = \(\frac{1}{√3}\) }

So,

θ = 60º

Now,

cot2θ

= cot ( 2 × 60º)

= cot 120º

= cot ( 180º - 60º )

{ cot is negative in 2nd quadrant }

= - cot120º

= - \(\frac{1}{√3}\)