The value of $$\cos^2 10° - \cos 10° \cos 50° + \cos^2 50°$$ is:

We have to find the value of the expression $$\cos^2 10^\circ-\cos 10^\circ\cos 50^\circ+\cos^2 50^\circ.$$

First, we convert each squared cosine term by using the identity

$$\cos^2\theta=\frac{1+\cos 2\theta}{2}.$$

Applying this to the two squared terms gives

$$\cos^2 10^\circ=\frac{1+\cos 20^\circ}{2},\qquad \cos^2 50^\circ=\frac{1+\cos 100^\circ}{2}.$$

Substituting these into the original expression, we obtain

$$\cos^2 10^\circ-\cos 10^\circ\cos 50^\circ+\cos^2 50^\circ =\frac{1+\cos 20^\circ}{2}-\cos 10^\circ\cos 50^\circ +\frac{1+\cos 100^\circ}{2}.$$

Now we combine the two half-fractions:

$$\frac{1+\cos 20^\circ}{2}+\frac{1+\cos 100^\circ}{2} =\frac{2+\cos 20^\circ+\cos 100^\circ}{2} =1+\frac{\cos 20^\circ+\cos 100^\circ}{2}.$$

So the whole expression turns into

$$1+\frac{\cos 20^\circ+\cos 100^\circ}{2}-\cos 10^\circ\cos 50^\circ.$$

Next we simplify $$\cos 100^\circ$$. Since $$100^\circ=180^\circ-80^\circ,$$ we use $$\cos(180^\circ-\alpha)=-\cos\alpha$$ to get

$$\cos 100^\circ=-\cos 80^\circ.$$

Hence $$\cos 20^\circ+\cos 100^\circ=\cos 20^\circ-\cos 80^\circ,$$ and the expression becomes

$$1+\frac{\cos 20^\circ-\cos 80^\circ}{2}-\cos 10^\circ\cos 50^\circ.$$

To handle the product $$\cos 10^\circ\cos 50^\circ,$$ we employ the product-to-sum identity

$$\cos A\cos B=\frac{\cos(A+B)+\cos(A-B)}{2}.$$

Taking $$A=10^\circ, B=50^\circ$$ gives

$$\cos 10^\circ\cos 50^\circ =\frac{\cos(10^\circ+50^\circ)+\cos(10^\circ-50^\circ)}{2} =\frac{\cos 60^\circ+\cos(-40^\circ)}{2}.$$

Because $$\cos(-\alpha)=\cos\alpha,$$ we have $$\cos(-40^\circ)=\cos 40^\circ,$$ and since $$\cos 60^\circ=\frac12,$$ we get

$$\cos 10^\circ\cos 50^\circ =\frac{\frac12+\cos 40^\circ}{2} =\frac14+\frac{\cos 40^\circ}{2}.$$

Substituting this back, we arrive at

$$1+\frac{\cos 20^\circ-\cos 80^\circ}{2}-\left(\frac14+\frac{\cos 40^\circ}{2}\right) =1-\frac14+\frac{\cos 20^\circ-\cos 80^\circ-\cos 40^\circ}{2}.$$

The constant terms combine to give $$\frac34,$$ so we now focus on the bracket:

$$\cos 20^\circ-\cos 80^\circ-\cos 40^\circ.$$

We first simplify $$\cos 20^\circ-\cos 40^\circ$$ using the difference identity

$$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}.$$

Here $$A=20^\circ, B=40^\circ,$$ so

$$\cos 20^\circ-\cos 40^\circ =-2\sin\frac{20^\circ+40^\circ}{2}\sin\frac{20^\circ-40^\circ}{2} =-2\sin 30^\circ\sin(-10^\circ).$$

Because $$\sin 30^\circ=\frac12$$ and $$\sin(-10^\circ)=-\sin 10^\circ,$$ we get

$$\cos 20^\circ-\cos 40^\circ =-2\left(\frac12\right)(-\sin 10^\circ) =\sin 10^\circ.$$

So the whole bracket becomes

$$\sin 10^\circ-\cos 80^\circ.$$

But $$\cos 80^\circ=\sin 10^\circ$$ (since $$\cos\theta=\sin(90^\circ-\theta)$$), therefore

$$\sin 10^\circ-\cos 80^\circ=\sin 10^\circ-\sin 10^\circ=0.$$

Thus the entire numerator of the fraction is zero, and the fractional part vanishes. We are left with

$$\frac34+0=\frac34.$$

Hence, the correct answer is Option A.