sin

(b) 2

We have,

$\sin ^{2}\left(\frac{\pi}{18}\right)+\sin ^{2}\left(\frac{\pi}{9}\right)+\sin ^{2}\left(\frac{7 \pi}{18}\right)+\sin ^{2}\left(\frac{4 \pi}{9}\right)$

$=\frac{1}{2}\left[1-\cos \left(\frac{\pi}{9}\right)+1-\cos \left(\frac{2 \pi}{9}\right)+1-\cos \frac{7 \pi}{9}+1-\cos \frac{8 \pi}{9}\right]$  $\left(\because \sin ^{2} \theta=\frac{1-\cos 2 \theta}{2}\right)$

$=\frac{1}{2}\left[4-\cos \left(\frac{\pi}{9}\right)-\cos \left(\frac{2 \pi}{9}\right)-\left\{-\cos \left(\pi-\frac{7 \pi}{9}\right)\right\}-\left\{-\cos \left(\pi-\frac{8 \pi}{9}\right)\right\}\right]$

$=\frac{1}{2}\left[4-\cos \left(\frac{\pi}{9}\right)-\cos \left(\frac{2 \pi}{9}\right)+\cos \left(\frac{2 \pi}{9}\right)+\cos \left(\frac{\pi}{9}\right)\right]$

$=\frac{4}{2}$

$=2$