If `cos theta = (2t)/(1+t^2)`, find the value of tan `theta`.

Answer

To find the value of \( \tan \theta \) given that \( \cos \theta = \frac{2t}{1+t^2} \), we can follow these steps:

### Step 1: Use the Pythagorean Identity
We know from trigonometry that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Given \( \cos \theta = \frac{2t}{1+t^2} \), we can find \( \sin^2 \theta \):
\[ \sin^2 \theta = 1 - \cos^2 \theta \]
Substituting the value of \( \cos \theta \):
\[ \sin^2 \theta = 1 - \left(\frac{2t}{1+t^2}\right)^2 \]

### Step 2: Simplify \( \sin^2 \theta \)
Calculating \( \cos^2 \theta \):
\[ \cos^2 \theta = \frac{4t^2}{(1+t^2)^2} \]
Now substituting this into the equation for \( \sin^2 \theta \):
\[ \sin^2 \theta = 1 - \frac{4t^2}{(1+t^2)^2} \]
To combine the terms, we rewrite 1 as:
\[ 1 = \frac{(1+t^2)^2}{(1+t^2)^2} \]
Thus,
\[ \sin^2 \theta = \frac{(1+t^2)^2 - 4t^2}{(1+t^2)^2} \]
Expanding the numerator:
\[ (1+t^2)^2 = 1 + 2t^2 + t^4 \]
Therefore,
\[ \sin^2 \theta = \frac{1 + 2t^2 + t^4 - 4t^2}{(1+t^2)^2} = \frac{1 - 2t^2 + t^4}{(1+t^2)^2} \]

### Step 3: Factor the Numerator
The numerator \( 1 - 2t^2 + t^4 \) can be factored as:
\[ (1 - t^2)^2 \]
Thus,
\[ \sin^2 \theta = \frac{(1 - t^2)^2}{(1+t^2)^2} \]

### Step 4: Find \( \sin \theta \)
Taking the square root to find \( \sin \theta \):
\[ \sin \theta = \frac{1 - t^2}{1 + t^2} \]

### Step 5: Calculate \( \tan \theta \)
Now we can find \( \tan \theta \) using the definition:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Substituting the values we found:
\[ \tan \theta = \frac{\frac{1 - t^2}{1 + t^2}}{\frac{2t}{1 + t^2}} = \frac{1 - t^2}{2t} \]

### Final Result
Thus, the value of \( \tan \theta \) is:
\[ \tan \theta = \frac{1 - t^2}{2t} \]