The value of `(tan^(2)20^(@)-sin^(2)20^(@))/(tan^(2)20^(@).sin^(2)20^(@))` is

Options

A. `1//2`
B. 1
C. 2
D. none of these

Explanation

To solve the expression \((\tan^2 20^\circ - \sin^2 20^\circ) / (\tan^2 20^\circ \cdot \sin^2 20^\circ)\), we will break it down step by step.

### Step 1: Rewrite the expression
We start with the given expression:
\[ \frac{\tan^2 20^\circ - \sin^2 20^\circ}{\tan^2 20^\circ \cdot \sin^2 20^\circ} \]

### Step 2: Separate the terms
We can separate the expression into two fractions:
\[ \frac{\tan^2 20^\circ}{\tan^2 20^\circ \cdot \sin^2 20^\circ} - \frac{\sin^2 20^\circ}{\tan^2 20^\circ \cdot \sin^2 20^\circ} \]

### Step 3: Simplify the first term
The first term simplifies as follows:
\[ \frac{\tan^2 20^\circ}{\tan^2 20^\circ \cdot \sin^2 20^\circ} = \frac{1}{\sin^2 20^\circ} \]
This is because \(\tan^2 20^\circ\) cancels out.

### Step 4: Simplify the second term
The second term simplifies as follows:
\[ \frac{\sin^2 20^\circ}{\tan^2 20^\circ \cdot \sin^2 20^\circ} = \frac{1}{\tan^2 20^\circ} \]
Again, \(\sin^2 20^\circ\) cancels out.

### Step 5: Combine the simplified terms
Now we can combine the two simplified terms:
\[ \frac{1}{\sin^2 20^\circ} - \frac{1}{\tan^2 20^\circ} \]

### Step 6: Rewrite \(\tan^2\) in terms of \(\sin\) and \(\cos\)
Recall that \(\tan^2 20^\circ = \frac{\sin^2 20^\circ}{\cos^2 20^\circ}\). Thus,
\[ \frac{1}{\tan^2 20^\circ} = \frac{\cos^2 20^\circ}{\sin^2 20^\circ} \]

### Step 7: Substitute back into the expression
Substituting this back gives us:
\[ \frac{1}{\sin^2 20^\circ} - \frac{\cos^2 20^\circ}{\sin^2 20^\circ} \]

### Step 8: Combine the fractions
Combining these fractions gives:
\[ \frac{1 - \cos^2 20^\circ}{\sin^2 20^\circ} \]

### Step 9: Use the Pythagorean identity
Using the identity \(1 - \cos^2 \theta = \sin^2 \theta\), we have:
\[ \frac{\sin^2 20^\circ}{\sin^2 20^\circ} = 1 \]

### Final Answer
Thus, the value of the given expression is:
\[ \boxed{1} \]