If `sectheta+tantheta=sqrt(3)(o^(@)lethetale90^(@))` then `tantheta` is

Options

A. underfined
B. `(1)/(sqrt(3))`
C. `(1)/(sqrt(2))`
D. `sqrt(3)`

Answer

To solve the equation \( \sec \theta + \tan \theta = \sqrt{3} \) for \( \tan \theta \), we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Thus, we can rewrite the equation as:
\[ \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \sqrt{3} \]
Combining the terms on the left side gives:
\[ \frac{1 + \sin \theta}{\cos \theta} = \sqrt{3} \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying yields:
\[ 1 + \sin \theta = \sqrt{3} \cos \theta \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ \sqrt{3} \cos \theta - \sin \theta = 1 \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides results in:
\[ (\sqrt{3} \cos \theta - \sin \theta)^2 = 1^2 \]
Expanding the left side:
\[ 3 \cos^2 \theta - 2\sqrt{3} \cos \theta \sin \theta + \sin^2 \theta = 1 \]

### Step 5: Use the Pythagorean identity
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we can substitute \( \sin^2 \theta = 1 - \cos^2 \theta \):
\[ 3 \cos^2 \theta - 2\sqrt{3} \cos \theta \sin \theta + (1 - \cos^2 \theta) = 1 \]
This simplifies to:
\[ 2 \cos^2 \theta - 2\sqrt{3} \cos \theta \sin \theta = 0 \]

### Step 6: Factor out the common term
Factoring out \( 2 \cos \theta \):
\[ 2 \cos \theta (\cos \theta - \sqrt{3} \sin \theta) = 0 \]
This gives us two cases:
1. \( \cos \theta = 0 \) (not valid in the first quadrant)
2. \( \cos \theta - \sqrt{3} \sin \theta = 0 \)

### Step 7: Solve for \( \tan \theta \)
From the second case:
\[ \cos \theta = \sqrt{3} \sin \theta \]
Dividing both sides by \( \cos \theta \) gives:
\[ 1 = \sqrt{3} \tan \theta \]
Thus:
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

### Final Answer
Therefore, the value of \( \tan \theta \) is:
\[ \tan \theta = \frac{1}{\sqrt{3}} \]