If `tanA+sinA=p and tanA-sinA=q`, then the value of `((p^(2)-q^(2))^(2))/(pq)` is :

Options

A. 16
B. 22
C. 18
D. 42

Answer

To solve the problem, we start with the equations given:

1. \( \tan A + \sin A = p \)
2. \( \tan A - \sin A = q \)

We need to find the value of \( \frac{(p^2 - q^2)^2}{pq} \).

### Step 1: Calculate \( p^2 - q^2 \)

Using the difference of squares formula, we have:

\[ p^2 - q^2 = (p + q)(p - q) \]

Now, we can find \( p + q \) and \( p - q \):

\[ p + q = (\tan A + \sin A) + (\tan A - \sin A) = 2\tan A \]

\[ p - q = (\tan A + \sin A) - (\tan A - \sin A) = 2\sin A \]

Thus, we can substitute these into our difference of squares:

\[ p^2 - q^2 = (2\tan A)(2\sin A) = 4\tan A \sin A \]

### Step 2: Calculate \( pq \)

Now, we calculate \( pq \):

\[ pq = (\tan A + \sin A)(\tan A - \sin A) = \tan^2 A - \sin^2 A \]

Using the identity \( \tan^2 A = \frac{\sin^2 A}{\cos^2 A} \):

\[ pq = \frac{\sin^2 A}{\cos^2 A} - \sin^2 A = \sin^2 A \left( \frac{1}{\cos^2 A} - 1 \right) = \sin^2 A \left( \frac{1 - \cos^2 A}{\cos^2 A} \right) \]

Since \( 1 - \cos^2 A = \sin^2 A \):

\[ pq = \sin^2 A \cdot \frac{\sin^2 A}{\cos^2 A} = \frac{\sin^4 A}{\cos^2 A} \]

### Step 3: Substitute into the original expression

Now we substitute \( p^2 - q^2 \) and \( pq \) into the expression we want to evaluate:

\[ \frac{(p^2 - q^2)^2}{pq} = \frac{(4\tan A \sin A)^2}{\frac{\sin^4 A}{\cos^2 A}} \]

Calculating \( (4\tan A \sin A)^2 \):

\[ (4\tan A \sin A)^2 = 16\tan^2 A \sin^2 A \]

Now substituting back:

\[ \frac{16\tan^2 A \sin^2 A}{\frac{\sin^4 A}{\cos^2 A}} = 16\tan^2 A \sin^2 A \cdot \frac{\cos^2 A}{\sin^4 A} = 16 \cdot \frac{\tan^2 A \cos^2 A}{\sin^2 A} \]

Since \( \tan^2 A = \frac{\sin^2 A}{\cos^2 A} \):

\[ = 16 \cdot \frac{\sin^2 A}{\cos^2 A} \cdot \frac{\cos^2 A}{\sin^2 A} = 16 \]

### Final Answer

Thus, the value of \( \frac{(p^2 - q^2)^2}{pq} \) is:

\[ \boxed{16} \]