Find the exact value of the expression. Do not use a calculator. 1 + sin^2 80^circ + sin^2 10^circ

Using the trigonometric identity: {eq}sin^2\theta + cos^2\theta = 1 {/eq}, we can rewrite the given expression as follows: {eq}1 + sin^2 80^circ + sin^2 10^circ = 1 + cos^2 10^circ + cos^2 80^circ {/eq} Now we can use another trigonometric identity: {eq}cos(theta+phi) = costhetacosphi - sinthetasinphi {/eq} We have: {eq}cos 80^circ = cos(50^circ+30^circ) {/eq} {eq} = cos 50^circ cos 30^circ - sin 50^circ sin 30^circ {/eq} {eq}= \frac{sqrt{3}}{2}cos 50^circ - \frac{1}{2}sin 50^circ {/eq} Similarly, {eq}cos 10^circ = cos(50^circ-40^circ) {/eq} {eq}=cos 50^circ cos 40^circ + sin 50^circ sin 40^circ {/eq} {eq}= \frac{sqrt{3}}{2}cos 50^circ + \frac{1}{2}sin 50^circ {/eq} Therefore, {eq}cos^2 80^circ = \frac{3}{4}cos^2 50^circ + \frac{1}{4}sin^2 50^circ - \frac{sqrt{3}}{2}cos 50^circsin 50^circ {/eq} And {eq}cos^2 10^circ = \frac{3}{4}cos^2 50^circ + \frac{1}{4}sin^2 50^circ + \frac{sqrt{3}}{2}cos 50^circsin 50^circ {/eq} Therefore, {eq}1 + sin^2 80^circ + sin^2 10^circ = \frac{3}{2}cos^2 50^circ+frac{1}{2}sin^2 50^circ{=}boxed{frac{3}{2}} {/eq}