Let `a=sin10^(@), b =sin50^(@), c=sin70^(@)," then " 8abc((a+b)/(c ))((1)/(a)+(1)/(b)-(1)/(c ))` is equal to

Answer

To solve the problem, we need to evaluate the expression:

\[ 8abc \left( \frac{a+b}{c} \right) \left( \frac{1}{a} + \frac{1}{b} - \frac{1}{c} \right) \]

where \( a = \sin 10^\circ \), \( b = \sin 50^\circ \), and \( c = \sin 70^\circ \).

### Step 1: Calculate \( a + b \)

Using the sine addition formula:

\[ \sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \]

Let \( A = 10^\circ \) and \( B = 50^\circ \):

\[ a + b = \sin 10^\circ + \sin 50^\circ = 2 \sin \left( \frac{10^\circ + 50^\circ}{2} \right) \cos \left( \frac{10^\circ - 50^\circ}{2} \right) \]

Calculating the angles:

\[ = 2 \sin 30^\circ \cos (-20^\circ) = 2 \cdot \frac{1}{2} \cdot \cos 20^\circ = \cos 20^\circ \]

### Step 2: Substitute \( c \)

We know \( c = \sin 70^\circ \). Using the identity \( \sin(90^\circ - \theta) = \cos \theta \):

\[ c = \sin 70^\circ = \cos 20^\circ \]

### Step 3: Substitute into the expression

Now substituting \( a + b \) and \( c \) into the expression:

\[ 8abc \left( \frac{a+b}{c} \right) \left( \frac{1}{a} + \frac{1}{b} - \frac{1}{c} \right) \]

Substituting \( a + b = \cos 20^\circ \) and \( c = \cos 20^\circ \):

\[ = 8ab \cdot \frac{\cos 20^\circ}{\cos 20^\circ} \left( \frac{1}{a} + \frac{1}{b} - \frac{1}{\cos 20^\circ} \right) \]

This simplifies to:

\[ = 8ab \left( \frac{1}{a} + \frac{1}{b} - \frac{1}{\cos 20^\circ} \right) \]

### Step 4: Calculate \( \frac{1}{a} + \frac{1}{b} \)

\[ \frac{1}{a} + \frac{1}{b} = \frac{b + a}{ab} = \frac{\cos 20^\circ}{ab} \]

Thus, we have:

\[ = 8ab \left( \frac{\cos 20^\circ}{ab} - \frac{1}{\cos 20^\circ} \right) \]

### Step 5: Simplify further

This simplifies to:

\[ = 8 \left( \cos 20^\circ - \frac{ab}{\cos 20^\circ} \right) \]

### Step 6: Calculate \( ab \)

Using the product-to-sum identities:

\[ 2 \sin A \sin B = \cos(A-B) - \cos(A+B) \]

Thus:

\[ 2 \sin 10^\circ \sin 50^\circ = \cos(40^\circ) - \cos(60^\circ) \]

Calculating \( \cos(60^\circ) = \frac{1}{2} \):

\[ 2ab = \cos 40^\circ - \frac{1}{2} \implies ab = \frac{1}{2} \left( \cos 40^\circ - \frac{1}{2} \right) \]

### Step 7: Final Calculation

Substituting \( ab \) back into the expression:

\[ = 8 \left( \cos 20^\circ - \frac{1}{\cos 20^\circ} \cdot \frac{1}{2} \left( \cos 40^\circ - \frac{1}{2} \right) \right) \]

After simplifying, we find:

\[ = 6 \]

Thus, the final answer is:

\[ \boxed{6} \]