If $$2(\cosec^2 39^\circ - \tan^2 51^\circ) - \frac{2}{3} \sin 90^\circ - \tan^2 56^\circ y \tan^2 34^\circ = \frac{y}{3}$$, then the value of $$y$$ is:

$$2(\cosec^2 39^\circ - \tan^2 51^\circ) - \frac{2}{3} \sin 90^\circ - \tan^2 56^\circ y \tan^2 34^\circ = \frac{y}{3}$$

$$=$$> $$2(\operatorname{cosec}^239^{\circ}-\tan^2\left(90-39\right)^{\circ})-\frac{2}{3}\left(1\right)-\tan^256^{\circ}y\tan^2\left(90-56\right)^{\circ}=\frac{y}{3}$$

$$=$$> $$2(\operatorname{cosec}^239^{\circ}-\cot^239)^{\circ\ }-\frac{2}{3}-\tan^256^{\circ}y\cot^256^{\circ}=\frac{y}{3}$$

$$=$$> $$2\left(1\right)-\frac{2}{3}-y=\frac{y}{3}$$

$$=$$> $$\frac{y}{3}+y=\frac{4}{3}$$

$$=$$> $$\frac{4y}{3}=\frac{4}{3}$$

$$=$$> $$y=1$$

Hence, the correct answer is Option C