To solve the given problem, we need to evaluate the expression:
\[\frac{\sin^6 15^\circ + \sin^6 75^\circ + 6 \sin^2 15^\circ + \sin^2 75^\circ} {\sin^4 15^\circ + \sin^4 75^\circ + 5 \sin^2 15^\circ \sin^2 75^\circ}\]Let's simplify this step by step:
Then, the expression becomes:
\[\frac{\sin^6 15^\circ + \cos^6 15^\circ + 6 \sin^2 15^\circ + \cos^2 15^\circ} {\sin^4 15^\circ + \cos^4 15^\circ + 5 \sin^2 15^\circ \cos^2 15^\circ}\]Thus, we can express:
\[\sin^6 15^\circ + \cos^6 15^\circ = (1)(\sin^4 15^\circ - \sin^2 15^\circ \cos^2 15^\circ + \cos^4 15^\circ) = \sin^4 15^\circ - \sin^2 15^\circ \cos^2 15^\circ + \cos^4 15^\circ\]Conclusion:
In both the numerator and the denominator, add the terms:
\[numerator = denominator = \sin^4 15^\circ + \cos^4 15^\circ + 5 \sin^2 15^\circ \cos^2 15^\circ + \sin^2 15^\circ \cos^2 15^\circ\]This simplifies to 1.
Therefore, the value of the expression is: 1
Step 1: Define variables.
Let \[ a = \sin^2 15^\circ, \quad b = \sin^2 75^\circ \]
But since \(\sin 75^\circ = \cos 15^\circ\), we get
\[ b = \cos^2 15^\circ \]
Therefore,
\[ a + b = \sin^2 15^\circ + \cos^2 15^\circ = 1 \]
Step 2: Write the given expression in terms of \(a\) and \(b\).
The expression is:
\[ \frac{\sin^6 15^\circ + \sin^6 75^\circ + 6 \sin^2 15^\circ \sin^2 75^\circ}{\sin^4 15^\circ + \sin^4 75^\circ + 5 \sin^2 15^\circ \sin^2 75^\circ} \]
Substitute \(a = \sin^2 15^\circ, \, b = \sin^2 75^\circ\):
\[ \frac{a^3 + b^3 + 6ab}{a^2 + b^2 + 5ab} \]
Step 3: Simplify the numerator.
Recall the identity:
\[ a^3 + b^3 = (a+b)(a^2 + b^2 - ab) \]
So, the numerator becomes:
\[ (a+b)(a^2 + b^2 - ab) + 6ab \]
Step 4: Simplify the denominator.
Since \[ a^2 + b^2 = (a+b)^2 - 2ab, \] we get
\[ a^2 + b^2 + 5ab = (a+b)^2 + 3ab \]
Step 5: Substitute \(a+b=1\).
Numerator:
\[ (1)\big((1)^2 - 3ab\big) + 6ab = 1 - 3ab + 6ab = 1 + 3ab \]
Denominator:
\[ (1)^2 + 3ab = 1 + 3ab \]
Step 6: Final simplification.
\[ \frac{1 + 3ab}{1 + 3ab} = 1 \]
\[ \boxed{1} \]
Option C