Discussion

Explanation:

$\frac{{\sin }^{6}15°+{\sin }^{6}75°+6{\sin }^{2}15°{\sin }^{2}75°}{{\sin }^{4}15°+{\sin }^{4}75°+5{\sin }^{2}15°{\sin }^{2}75°}$

Let sin²15° = a and sin²75° = b

∴ $\frac{{\sin }^{6}15°+{\sin }^{6}75°+6{\sin }^{2}15°{\sin }^{2}75°}{{\sin }^{4}15°+{\sin }^{4}75°+5{\sin }^{2}15°{\sin }^{2}75°}$ = $\frac{{\mathrm{a}}^3+{\mathrm{b}}^3+6\mathrm{ab}}{{\mathrm{a}}^2+{\mathrm{b}}^2+5\mathrm{ab}}$

Now, a³ + b³ = (a + b)³ - 3ab(a + b), and
a² + b² = (a + b)² - 2ab

∴ $\frac{{\sin }^{6}15°+{\sin }^{6}75°+6{\sin }^{2}15°{\sin }^{2}75°}{{\sin }^{4}15°+{\sin }^{4}75°+5{\sin }^{2}15°{\sin }^{2}75°}$ = $\frac{{(\mathrm{a}+\mathrm{b})}^3-3\mathrm{ab}(\mathrm{a}+\mathrm{b})+6\mathrm{ab}}{{(\mathrm{a}+\mathrm{b})}^2-2\mathrm{ab}+5\mathrm{ab}}$

Now, a + b = sin²15° + sin²75° = sin²15° + cos²15° = 1 [∵ Sin𝜃 = Cos(90 - 𝜃)]

∴ $\frac{{\sin }^{6}15°+{\sin }^{6}75°+6{\sin }^{2}15°{\sin }^{2}75°}{{\sin }^{4}15°+{\sin }^{4}75°+5{\sin }^{2}15°{\sin }^{2}75°}$ = $\frac{1-3\mathrm{ab}+6\mathrm{ab}}{1-2\mathrm{ab}+5\mathrm{ab}}$ = $\frac{1+3\mathrm{ab}}{1+3\mathrm{ab}}$ = 1

Hence, option (c). 

» Your doubt will be displayed only after approval.