The value of sin²30 – sin²40 + sin²45 – sin²55 – sin²35 + sin²45 – sin²50 + sin²60 is:

To find the value of the given expression, we can write it down clearly:
$E={\sin }^{2}30°-{\sin }^{2}40°+{\sin }^{2}45°-{\sin }^{2}55°-{\sin }^{2}35°+{\sin }^{2}45°-{\sin }^{2}50°+{\sin }^{2}60°$

We can group the terms to simplify the calculation:
$E=({\sin }^{2}30°+{\sin }^{2}60°)+({\sin }^{2}45°+{\sin }^{2}45��)-({\sin }^{2}40°+{\sin }^{2}50°)-({\sin }^{2}35°+{\sin }^{2}55°)$

Now, let's simplify each grouped pair using trigonometric properties:

1. First Pair:
We know the standard values of $\sin 30°=\frac{1}{2}$ and $\sin 60°=\frac{\sqrt{3}}{2}$.
Therefore:
${\sin }^{2}30°+{\sin }^{2}60°={\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{4}+\frac{3}{4}=1$

2. Second Pair:
We know the standard value of $\sin 45°=\frac{1}{\sqrt{2}}$.
Therefore:
${\sin }^{2}45°+{\sin }^{2}45°={(\frac{1}{\sqrt{2}}}^2+{(\frac{1}{\sqrt{2}}}^2=\frac{1}{2}+\frac{1}{2}=1$

3. Third Pair:
Using the complementary angle formula, $\sin (90°-\theta )=\cos \theta$:
$\sin 50°=\sin (90°-40°)=\cos 40°$
Thus, using the identity ${\sin }^2\theta +{\cos }^2\theta =1$:
${\sin }^{2}40°+{\sin }^{2}50°={\sin }^{2}40°+{\cos }^{2}40°=1$

4. Fourth Pair:
Similarly, for the complementary angle of $55°$:
$\sin 55°=\sin (90°-35°)=\cos 35°$
Thus:
${\sin }^{2}35°+{\sin }^{2}55°={\sin }^{2}35°+{\cos }^{2}35°=1$

Substitute these values back into the grouped expression:
$E=1+1-1-1=0$

Thus, the value of the given expression is 0.