If `sec alpha = (5)/(4)`, then `(tan alpha)/(1+tan^(2)alpha)` is equal to

Options

A. `(9)/(25)`
B. `(12)/(25)`
C. `(3)/(4)`
D. `(1)/(25)`

Answer

To solve the problem, we need to find the value of \(\frac{\tan \alpha}{1 + \tan^2 \alpha}\) given that \(\sec \alpha = \frac{5}{4}\).

### Step 1: Use the identity relating secant and tangent
We know the identity:
\[ \sec^2 \alpha = 1 + \tan^2 \alpha \]
From the problem, we have:
\[ \sec \alpha = \frac{5}{4} \]
So, we can find \(\sec^2 \alpha\):
\[ \sec^2 \alpha = \left(\frac{5}{4}\right)^2 = \frac{25}{16} \]

### Step 2: Substitute \(\sec^2 \alpha\) into the identity
Using the identity:
\[ \sec^2 \alpha = 1 + \tan^2 \alpha \]
we can substitute \(\sec^2 \alpha\):
\[ \frac{25}{16} = 1 + \tan^2 \alpha \]

### Step 3: Solve for \(\tan^2 \alpha\)
Rearranging the equation gives:
\[ \tan^2 \alpha = \frac{25}{16} - 1 \]
To subtract 1, we express 1 as \(\frac{16}{16}\):
\[ \tan^2 \alpha = \frac{25}{16} - \frac{16}{16} = \frac{25 - 16}{16} = \frac{9}{16} \]

### Step 4: Find \(\tan \alpha\)
Taking the square root of both sides, we find:
\[ \tan \alpha = \sqrt{\frac{9}{16}} = \frac{3}{4} \]

### Step 5: Calculate \(1 + \tan^2 \alpha\)
Now we need to calculate \(1 + \tan^2 \alpha\):
\[ 1 + \tan^2 \alpha = 1 + \frac{9}{16} = \frac{16}{16} + \frac{9}{16} = \frac{25}{16} \]

### Step 6: Substitute into the expression
Now we can substitute \(\tan \alpha\) and \(1 + \tan^2 \alpha\) into the expression:
\[ \frac{\tan \alpha}{1 + \tan^2 \alpha} = \frac{\frac{3}{4}}{\frac{25}{16}} \]

### Step 7: Simplify the expression
To simplify, we multiply by the reciprocal:
\[ \frac{3}{4} \times \frac{16}{25} = \frac{3 \times 16}{4 \times 25} = \frac{48}{100} = \frac{12}{25} \]

### Final Answer
Thus, the value of \(\frac{\tan \alpha}{1 + \tan^2 \alpha}\) is:
\[ \frac{12}{25} \]

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