If `(tantheta+cottheta)/(tantheta-cottheta)=2,(0^(@)lethetale90^(@))`, then the value of `sintheta` is

Options

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

Answer

To solve the equation \(\frac{\tan \theta + \cot \theta}{\tan \theta - \cot \theta} = 2\), we will follow these steps:

### Step 1: Cross-multiply the equation
Starting from the equation:
\[ \frac{\tan \theta + \cot \theta}{\tan \theta - \cot \theta} = 2 \]
Cross-multiplying gives us:
\[ \tan \theta + \cot \theta = 2(\tan \theta - \cot \theta) \]

### Step 2: Expand and rearrange the equation
Expanding the right side:
\[ \tan \theta + \cot \theta = 2\tan \theta - 2\cot \theta \]
Now, rearranging the terms:
\[ \tan \theta + \cot \theta - 2\tan \theta + 2\cot \theta = 0 \]
This simplifies to:
\[ -\tan \theta + 3\cot \theta = 0 \]

### Step 3: Isolate \(\cot \theta\)
Rearranging gives:
\[ 3\cot \theta = \tan \theta \]
Dividing both sides by \(\tan \theta\):
\[ \frac{\cot \theta}{\tan \theta} = \frac{1}{3} \]
Since \(\cot \theta = \frac{1}{\tan \theta}\), we can write:
\[ \frac{1}{\tan^2 \theta} = \frac{1}{3} \]

### Step 4: Solve for \(\tan^2 \theta\)
Taking the reciprocal gives:
\[ \tan^2 \theta = 3 \]

### Step 5: Find \(\tan \theta\)
Taking the square root:
\[ \tan \theta = \sqrt{3} \]
Since \(\theta\) is in the range \(0^\circ < \theta < 90^\circ\), we have:
\[ \tan \theta = \sqrt{3} \]

### Step 6: Determine the angle \(\theta\)
The angle \(\theta\) for which \(\tan \theta = \sqrt{3}\) is:
\[ \theta = 60^\circ \]

### Step 7: Find \(\sin \theta\)
Now, we need to find \(\sin \theta\):
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

### Final Answer
Thus, the value of \(\sin \theta\) is:
\[ \frac{\sqrt{3}}{2} \]

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