If sin `theta=7/25` and `90^@ lt theta lt 180^@`, find the value of sec `theta+tan theta`.

Answer

To find the value of \( \sec \theta + \tan \theta \) given that \( \sin \theta = \frac{7}{25} \) and \( 90^\circ < \theta < 180^\circ \), we can follow these steps:

### Step 1: Identify the quadrant
Since \( \theta \) is between \( 90^\circ \) and \( 180^\circ \), it lies in the second quadrant. In this quadrant, sine is positive, while cosine and tangent are negative.

### Step 2: Use the Pythagorean identity
We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting the value of \( \sin \theta \):
\[ \left(\frac{7}{25}\right)^2 + \cos^2 \theta = 1 \]
Calculating \( \sin^2 \theta \):
\[ \frac{49}{625} + \cos^2 \theta = 1 \]
Subtracting \( \frac{49}{625} \) from both sides:
\[ \cos^2 \theta = 1 - \frac{49}{625} = \frac{625 - 49}{625} = \frac{576}{625} \]

### Step 3: Find \( \cos \theta \)
Taking the square root:
\[ \cos \theta = -\sqrt{\frac{576}{625}} = -\frac{24}{25} \]
(Note: We take the negative root because cosine is negative in the second quadrant.)

### Step 4: Find \( \sec \theta \)
The secant function is the reciprocal of cosine:
\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{24}{25}} = -\frac{25}{24} \]

### Step 5: Find \( \tan \theta \)
The tangent function is given by:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{7}{25}}{-\frac{24}{25}} = -\frac{7}{24} \]

### Step 6: Calculate \( \sec \theta + \tan \theta \)
Now we can find \( \sec \theta + \tan \theta \):
\[ \sec \theta + \tan \theta = -\frac{25}{24} - \frac{7}{24} = -\frac{25 + 7}{24} = -\frac{32}{24} = -\frac{4}{3} \]

### Final Answer
Thus, the value of \( \sec \theta + \tan \theta \) is:
\[ \boxed{-\frac{4}{3}} \]