`(cos 12^(@) - sin 12^(@))/(cos 12^(@) + sin 12) + (sin 147^(@))/(cos 147^(@))=`

Options

A. 1
B. `-1`
C. 0
D. `sqrt3`

Answer

To solve the expression

\[ \frac{\cos 12^\circ - \sin 12^\circ}{\cos 12^\circ + \sin 12^\circ} + \frac{\sin 147^\circ}{\cos 147^\circ} \]

we will simplify it step by step.

### Step 1: Simplify the second term
We know that \( \sin 147^\circ \) can be expressed using the sine of a complementary angle:
\[ \sin 147^\circ = \sin(180^\circ - 33^\circ) = \sin 33^\circ \]
and
\[ \cos 147^\circ = \cos(180^\circ - 33^\circ) = -\cos 33^\circ \]
Thus, the second term becomes:
\[ \frac{\sin 147^\circ}{\cos 147^\circ} = \frac{\sin 33^\circ}{-\cos 33^\circ} = -\tan 33^\circ \]

### Step 2: Rewrite the expression
Now, we can rewrite the original expression as:
\[ \frac{\cos 12^\circ - \sin 12^\circ}{\cos 12^\circ + \sin 12^\circ} - \tan 33^\circ \]

### Step 3: Simplify the first term
To simplify the first term, we can factor out \( \cos 12^\circ \) from the numerator and denominator:
\[ = \frac{\cos 12^\circ(1 - \tan 12^\circ)}{\cos 12^\circ(1 + \tan 12^\circ)} = \frac{1 - \tan 12^\circ}{1 + \tan 12^\circ} \]

### Step 4: Use the tangent subtraction formula
We can use the tangent subtraction formula:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
Let \( A = 45^\circ \) and \( B = 12^\circ \):
\[ \tan(45^\circ - 12^\circ) = \frac{1 - \tan 12^\circ}{1 + \tan 12^\circ} \]
Thus, we have:
\[ \frac{1 - \tan 12^\circ}{1 + \tan 12^\circ} = \tan(45^\circ - 12^\circ) = \tan 33^\circ \]

### Step 5: Combine the terms
Now substituting back into our expression, we have:
\[ \tan 33^\circ - \tan 33^\circ = 0 \]

### Final Answer
Thus, the value of the entire expression is:
\[ \boxed{0} \]