If `A= tan27theta - tantheta` and `B= (sintheta)/(cos 3theta) +(sin 3theta)/(cos 9theta) + (sin 9theta)/(cos 27theta)`, then the value of `(A+B)/(A-B) +(A-B)/(A+B) -10/3` is

Explanation

To solve the given problem, we need to evaluate the expression \((A+B)/(A-B) +(A-B)/(A+B) -10/3\) where \(A = \tan(27\theta) - \tan(\theta)\) and \(B = \frac{\sin(\theta)}{\cos(3\theta)} + \frac{\sin(3\theta)}{\cos(9\theta)} + \frac{\sin(9\theta)}{\cos(27\theta)}\).

### Step 1: Simplify \(B\)

We start with the expression for \(B\):

\[ B = \frac{\sin(\theta)}{\cos(3\theta)} + \frac{\sin(3\theta)}{\cos(9\theta)} + \frac{\sin(9\theta)}{\cos(27\theta)} \]

We can rewrite each term in \(B\) using the identity \(\frac{\sin(x)}{\cos(y)} = \tan(x) \cdot \sec(y)\):

\[ B = \tan(\theta) \sec(3\theta) + \tan(3\theta) \sec(9\theta) + \tan(9\theta) \sec(27\theta) \]

### Step 2: Use the identity for \(B\)

We can express \(B\) in a different way. Notice that:

\[ B = \frac{\sin(\theta)}{\cos(3\theta)} + \frac{\sin(3\theta)}{\cos(9\theta)} + \frac{\sin(9\theta)}{\cos(27\theta)} \]

By using the double angle formula, we can rewrite:

\[ \frac{\sin(2\theta)}{\cos(3\theta)} = 2 \frac{\sin(\theta) \cos(\theta)}{\cos(3\theta)} \]

Continuing this process for each term, we can find a pattern or relation among the terms.

### Step 3: Find a relation between \(A\) and \(B\)

From the analysis, we find that:

\[ B = \frac{1}{2} A \]

This means:

\[ A = 2B \]

### Step 4: Substitute \(A\) and \(B\) into the expression

Now we substitute \(A\) and \(B\) into the expression we need to evaluate:

\[ \frac{A+B}{A-B} + \frac{A-B}{A+B} - \frac{10}{3} \]

Substituting \(A = 2B\):

\[ \frac{2B + B}{2B - B} + \frac{2B - B}{2B + B} - \frac{10}{3} \]

This simplifies to:

\[ \frac{3B}{B} + \frac{B}{3B} - \frac{10}{3} \]

### Step 5: Simplify the fractions

The expression simplifies to:

\[ 3 + \frac{1}{3} - \frac{10}{3} \]

### Step 6: Combine the terms

Now, we combine the terms:

\[ 3 = \frac{9}{3} \]

So,

\[ \frac{9}{3} + \frac{1}{3} - \frac{10}{3} = \frac{9 + 1 - 10}{3} = \frac{0}{3} = 0 \]

### Final Answer

Thus, the value of the expression is:

\[ \boxed{0} \]