If $\sin ^6 \theta+\cos ^6 \theta=\frac{1}{3}, 0^{\circ}<\theta<90^{\circ}$, then what is the value of $\sin \theta \cos \theta$ ?

$\frac{\sqrt{2}}{3}$

sin6 θ + cos6 θ = \(\frac{1}{3}\)

(sin² θ)³ + (cos² θ)³ = \(\frac{1}{3}\)

( sin² θ + cos² θ ) . ( sin⁴ θ + cos4 θ -  sinθ.cosθ ) = \(\frac{1}{3}\)

{ using , sin² θ + cos² θ = 1 }

( sin⁴ θ + cos4 θ -  sin²θ.cos²θ ) = \(\frac{1}{3}\)

( sin² θ + cos² θ ) -  sin²θ.cos²θ  = \(\frac{1}{3}\)

( sin² θ + cos² θ ) - 3 sin²θ.cos²θ  = \(\frac{1}{3}\)

1 - 3 sin²θ.cos²θ  = \(\frac{1}{3}\)

1 - \(\frac{1}{3}\) = 3 sin²θ.cos²θ

\(\frac{2}{9}\) =  sin²θ.cos²θ

sinθ.cosθ = \(\frac{√2}{3}\)