If sin² θ cos² θ = $\frac{2}{9}$, then what will be the value of cosec² θ + sec²θ ?

9/2

We are given that :-

sin²θ . cos²θ = \(\frac{2}{9}\)

Now,

cosec²θ + sec²θ

= \(\frac{1}{sin²θ}\) + \(\frac{1}{cos²θ}\)

= \(\frac{cos²θ + sin²θ}{sin²θ .cos²θ }\)

{ we know, sin²θ + cos²θ = 1 }

=  \(\frac{1}{sin²θ .cos²θ }\) 

= \(\frac{9}{2}\)    [ given :- sin²θ . cos²θ = \(\frac{2}{9}\) ]