Question

Find the value of  $$ \dfrac{\sin^6 15^\circ + \sin^6 75^\circ + 6 \sin^2 15^\circ + \sin^2 75^\circ}{\sin^4 15^\circ + \sin^4 75^\circ + 5 \sin^2 15^\circ \sin^2 75^\circ} $$ 

• A. sin 15°+sin 75°
• B. 6/5
• C. 1
• D. sin 15°+cos 15°
• E. None of the above

Answer 1

The Correct Option is C

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.

\[ \frac{1 + 3ab}{1 + 3ab} = 1 \]

Final Answer:

\[ \boxed{1} \]

Correct Option:

Option C