The value of `(cos 15^(@) - sin 15^(@))/(cos 15^(@) + sin 15^(@)` is

Options

A. 1
B. `sqrt(3)`
C. 0
D. `(1)/(sqrt(3))`

Answer

To solve the expression \(\frac{\cos 15^\circ - \sin 15^\circ}{\cos 15^\circ + \sin 15^\circ}\), we can follow these steps:

### Step 1: Multiply by the Conjugate
Multiply the numerator and the denominator by the conjugate of the denominator, which is \(\cos 15^\circ - \sin 15^\circ\).

\[ \frac{(\cos 15^\circ - \sin 15^\circ)(\cos 15^\circ - \sin 15^\circ)}{(\cos 15^\circ + \sin 15^\circ)(\cos 15^\circ - \sin 15^\circ)} \]

### Step 2: Simplify the Denominator
Using the difference of squares formula, we can simplify the denominator:

\[ (\cos 15^\circ + \sin 15^\circ)(\cos 15^\circ - \sin 15^\circ) = \cos^2 15^\circ - \sin^2 15^\circ \]

### Step 3: Use the Cosine Double Angle Identity
Recall the identity \(\cos^2 \theta - \sin^2 \theta = \cos 2\theta\). Therefore, we have:

\[ \cos^2 15^\circ - \sin^2 15^\circ = \cos 30^\circ \]

### Step 4: Simplify the Numerator
Now, simplify the numerator using the square of a binomial:

\[ (\cos 15^\circ - \sin 15^\circ)^2 = \cos^2 15^\circ - 2\cos 15^\circ \sin 15^\circ + \sin^2 15^\circ \]

Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\), we can write:

\[ \cos^2 15^\circ - 2\cos 15^\circ \sin 15^\circ + \sin^2 15^\circ = 1 - 2\cos 15^\circ \sin 15^\circ \]

### Step 5: Use the Sine Double Angle Identity
Recall that \(2\cos \theta \sin \theta = \sin 2\theta\). Thus, we can rewrite the numerator:

\[ 1 - \sin 30^\circ = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 6: Substitute Back into the Expression
Now substituting back into our expression, we have:

\[ \frac{\frac{1}{2}}{\cos 30^\circ} \]

### Step 7: Substitute the Value of \(\cos 30^\circ\)
We know that \(\cos 30^\circ = \frac{\sqrt{3}}{2}\). Therefore, substituting this value gives:

\[ \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \]

### Final Answer
Thus, the value of \(\frac{\cos 15^\circ - \sin 15^\circ}{\cos 15^\circ + \sin 15^\circ}\) is:

\[ \frac{1}{\sqrt{3}} \]

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