If `-tan theta = (sin13^@ + cos 13^@)/(sin13^@ - cos13^@)` , find `theta` ?

Options

A. `58^@`
B. `-58^@`
C. `63^@`
D. `-63^@`

Answer

To solve the equation \(-\tan \theta = \frac{\sin 13^\circ + \cos 13^\circ}{\sin 13^\circ - \cos 13^\circ}\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ -\tan \theta = \frac{\sin 13^\circ + \cos 13^\circ}{\sin 13^\circ - \cos 13^\circ} \]

### Step 2: Divide numerator and denominator by \(\cos 13^\circ\)
To simplify the right-hand side, we divide both the numerator and the denominator by \(\cos 13^\circ\):
\[ -\tan \theta = \frac{\frac{\sin 13^\circ}{\cos 13^\circ} + \frac{\cos 13^\circ}{\cos 13^\circ}}{\frac{\sin 13^\circ}{\cos 13^\circ} - \frac{\cos 13^\circ}{\cos 13^\circ}} \]
This simplifies to:
\[ -\tan \theta = \frac{\tan 13^\circ + 1}{\tan 13^\circ - 1} \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ -\tan \theta (\tan 13^\circ - 1) = \tan 13^\circ + 1 \]
Expanding this, we have:
\[ -\tan \theta \tan 13^\circ + \tan \theta = \tan 13^\circ + 1 \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:
\[ \tan \theta + \tan \theta \tan 13^\circ = \tan 13^\circ + 1 \]
Factoring out \(\tan \theta\) on the left side:
\[ \tan \theta (1 + \tan 13^\circ) = \tan 13^\circ + 1 \]

### Step 5: Solve for \(\tan \theta\)
Now, we can isolate \(\tan \theta\):
\[ \tan \theta = \frac{\tan 13^\circ + 1}{1 + \tan 13^\circ} \]
This simplifies to:
\[ \tan \theta = \tan(45^\circ + 13^\circ) \]
Thus, we have:
\[ \tan \theta = \tan 58^\circ \]

### Step 6: Finding \(\theta\)
Since \(\tan \theta = \tan 58^\circ\), we can conclude:
\[ \theta = 58^\circ + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]
For the principal value, we take:
\[ \theta = 58^\circ \]

### Final Answer
\(\theta = 58^\circ\)

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