Answer 1

\[\frac{2\pi}{5} = 72°, \frac{\pi}{3} = 60° \]

\[LHS = \sin^2 72° - \sin^2 6°\]

\[ = \sin^2 \left( 90° - 18° \right) - \frac{3}{4}\]

\[ = \cos^2 18 °- \frac{3}{4} \left( \because \sin\left( 90°  - \theta \right) = cos\theta \right)\]

\[ = \left( \frac{\sqrt{10 + 2\sqrt{5}}}{4} \right)^2 - \frac{3}{4} \left( \because \cos18°  = \frac{\sqrt{10 + 2\sqrt{5}}}{4} \right)\]

\[ = \frac{10 + 2\sqrt{5}}{16} - \frac{3}{4}\]

\[= \frac{10 + 2\sqrt{5} - 12}{16}\]
\[ = \frac{2\sqrt{5} - 2}{16}\]
\[ = \frac{\sqrt{5} - 1}{8}\]
\[ = RHS\]
\[\text{ Hence proved } .\]