If `sin20^(@)=p, " then find the value of " ((sin380^(@)-sin340^(@))/(cos380^(@)+cos340^(@)))`.

Options

A. `sqrt(1-P^(2))`
B. `sqrt((1-p^(2))/(p))`
C. `(p)/sqrt(1-p^(2))`
D. None of these

Explanation

To solve the problem, we need to find the value of the expression:

\[ \frac{\sin 380^\circ - \sin 340^\circ}{\cos 380^\circ + \cos 340^\circ} \]

Given that \(\sin 20^\circ = p\).

### Step 1: Simplify the angles using periodic properties
We can simplify \(\sin 380^\circ\) and \(\sin 340^\circ\) using the periodic properties of sine and cosine:

\[ \sin 380^\circ = \sin(360^\circ + 20^\circ) = \sin 20^\circ = p \]

\[ \sin 340^\circ = \sin(360^\circ - 20^\circ) = -\sin 20^\circ = -p \]

### Step 2: Substitute the values into the expression
Now we can substitute these values into the expression:

\[ \sin 380^\circ - \sin 340^\circ = p - (-p) = p + p = 2p \]

### Step 3: Simplify the cosine terms
Now, we simplify \(\cos 380^\circ\) and \(\cos 340^\circ\):

\[ \cos 380^\circ = \cos(360^\circ + 20^\circ) = \cos 20^\circ \]

\[ \cos 340^\circ = \cos(360^\circ - 20^\circ) = \cos 20^\circ \]

Thus,

\[ \cos 380^\circ + \cos 340^\circ = \cos 20^\circ + \cos 20^\circ = 2\cos 20^\circ \]

### Step 4: Substitute back into the expression
Now we can substitute back into the expression:

\[ \frac{\sin 380^\circ - \sin 340^\circ}{\cos 380^\circ + \cos 340^\circ} = \frac{2p}{2\cos 20^\circ} \]

### Step 5: Simplify the expression
The \(2\) cancels out:

\[ \frac{2p}{2\cos 20^\circ} = \frac{p}{\cos 20^\circ} \]

### Step 6: Express \(\cos 20^\circ\) in terms of \(p\)
Using the Pythagorean identity:

\[ \cos^2 20^\circ + \sin^2 20^\circ = 1 \]

Substituting \(\sin 20^\circ = p\):

\[ \cos^2 20^\circ + p^2 = 1 \implies \cos^2 20^\circ = 1 - p^2 \implies \cos 20^\circ = \sqrt{1 - p^2} \]

### Step 7: Final expression
Now substituting \(\cos 20^\circ\) back into our expression:

\[ \frac{p}{\cos 20^\circ} = \frac{p}{\sqrt{1 - p^2}} \]

Thus, the final answer is:

\[ \frac{p}{\sqrt{1 - p^2}} \]