Q. Evaluate : $\frac{7{\sin }^{2}30°+3 \cos e{c}^{2}60°- co{t}^{2}45°}{{\sin }^{2}70°+ {\sin }^{2}20°}+2 {\sin }^{2}30° - 2 {\cos }^{2}30°$

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$$\begin{gathered} We have: \frac{7{\sin }^{2}30° + 3\cos e{c}^{2}60° - co{t}^{2}45°}{{\sin }^{2}70° + {\sin }^{2}20°} + 2{\sin }^{2}30° - 2{\cos }^{2}30° \\ = \frac{7{\sin }^{2}30° + 3\cos e{c}^{2}60° - co{t}^{2}45°}{{\sin }^{2}70° + {\left(\sin \left(90 - 70\right)°\right)}^2} + 2{\sin }^{2}30° - 2{\cos }^{2}30° \\ = \frac{7{\sin }^{2}30° + 3\cos e{c}^{2}60° - co{t}^{2}45°}{{\sin }^{2}70° + {\cos }^{2}70°} + 2{\sin }^{2}30° - 2{\cos }^{2}30° \left[\because \sin \left(90° - x\right) = \cos x\right] \\ = \frac{7{\sin }^{2}30° + 3\cos e{c}^{2}60° - co{t}^{2}45°}{1} + 2{\sin }^{2}30° - 2{\cos }^{2}30° \left[\because {\sin }^{2}x + {\cos }^{2}x = 1\right] \\ = 7{\sin }^{2}30° + 3\cos e{c}^{2}60° - co{t}^{2}45° + 2{\sin }^{2}30° - 2{\cos }^{2}30° \\ = 7{\left(\frac{1}{2}\right)}^2 + 3{\left(\frac{2}{\sqrt{3}}\right)}^2 - {\left(1\right)}^2 + 2{\left(\frac{1}{2}\right)}^2 - 2{\left(\frac{\sqrt{3}}{2}\right)}^2 \\ = \frac{7}{4} + \left(3 \times \frac{4}{3}\right) - 1 + \left(2 \times \frac{1}{4}\right) - \left(2 \times \frac{3}{4}\right) \\ = \frac{7}{4} + 4 - 1 + \frac{1}{2} - \frac{3}{2} \\ = \frac{7}{4} + 2 = \boxed{\frac{\mathbf{15}}{\mathbf{4}}} \end{gathered}$$

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