Solve this

$\frac{\left(\sin ^{2} 23^{\circ}+\sin ^{2} 67^{\circ}\right)}{\left(\cos ^{2} 13^{\circ}+\cos ^{2} 77^{\circ}\right)}+\sin ^{2} 59^{\circ}+\cos 59^{\circ} \sin 31^{\circ}$

$=\left[\frac{\sin ^{2} 23^{*}+\left(\sin \left(90^{\circ}-23^{\circ}\right)\right)^{2}}{\cos ^{2} 13^{*}+\left(\cos \left(90^{\circ}-13^{\circ}\right)\right)^{2}}\right]+\sin ^{2} 59^{\circ}+\cos 59^{\circ} \sin \left(90^{\circ}-59^{\circ}\right)$

$=\left[\frac{\sin ^{2} 23^{\circ}+\cos ^{2} 23^{\circ}}{\cos ^{2} 13^{\circ}+\sin ^{2} 13^{\circ}}\right]+\sin ^{2} 59^{\circ}+\cos 59^{\circ} \cos 59^{\circ} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right.$ and $\left.\cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$

$=\left[\frac{1}{1}\right]+\sin ^{2} 59^{\circ}+\cos ^{2} 59^{\circ} \quad$ (using the identity : $\sin ^{2} \theta+\cos ^{2} \theta=1$ )

$=1+1$

$=2$

Hence, the correct option is (b).

Disclaimer: There must be $\sin 31^{\circ}$ instead of $\sin 21^{\circ}$.