If `tantheta+sectheta=1. 5 ,` find `sintheta,tanthetaa n dsecthetadot`

Answer

To solve the problem where \( \tan \theta + \sec \theta = 1.5 \), we will follow these steps:

### Step 1: Rewrite the equation
We can express \( 1.5 \) as a fraction:
\[ \tan \theta + \sec \theta = \frac{3}{2} \]

### Step 2: Rationalize the equation
To rationalize, we multiply both sides by \( \sec \theta - \tan \theta \):
\[ (\tan \theta + \sec \theta)(\sec \theta - \tan \theta) = \frac{3}{2}(\sec \theta - \tan \theta) \]
This gives us:
\[ \sec^2 \theta - \tan^2 \theta = \frac{3}{2}(\sec \theta - \tan \theta) \]

### Step 3: Use the Pythagorean identity
We know that:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
So we can substitute this into our equation:
\[ 1 = \frac{3}{2}(\sec \theta - \tan \theta) \]

### Step 4: Solve for \( \sec \theta - \tan \theta \)
Rearranging gives:
\[ \sec \theta - \tan \theta = \frac{2}{3} \]

### Step 5: Set up equations
Now we have two equations:
1. \( \sec \theta + \tan \theta = \frac{3}{2} \) (Equation 1)
2. \( \sec \theta - \tan \theta = \frac{2}{3} \) (Equation 2)

### Step 6: Add the equations
Adding Equation 1 and Equation 2:
\[ (\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) = \frac{3}{2} + \frac{2}{3} \]
This simplifies to:
\[ 2 \sec \theta = \frac{3}{2} + \frac{2}{3} \]

### Step 7: Find a common denominator
To add the fractions, we find a common denominator (which is 6):
\[ \frac{3}{2} = \frac{9}{6}, \quad \frac{2}{3} = \frac{4}{6} \]
Thus,
\[ 2 \sec \theta = \frac{9}{6} + \frac{4}{6} = \frac{13}{6} \]

### Step 8: Solve for \( \sec \theta \)
Dividing both sides by 2:
\[ \sec \theta = \frac{13}{12} \]

### Step 9: Find \( \sin \theta \) and \( \tan \theta \)
Using the relationship of secant:
\[ \sec \theta = \frac{1}{\cos \theta} \implies \cos \theta = \frac{12}{13} \]
Now to find \( \sin \theta \):
Using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \implies \sin^2 \theta + \left(\frac{12}{13}\right)^2 = 1 \]
Calculating \( \left(\frac{12}{13}\right)^2 \):
\[ \sin^2 \theta + \frac{144}{169} = 1 \implies \sin^2 \theta = 1 - \frac{144}{169} = \frac{25}{169} \]
Thus,
\[ \sin \theta = \frac{5}{13} \]

### Step 10: Find \( \tan \theta \)
Using the definition of tangent:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \]

### Final Results
Thus, we have:
\[ \sin \theta = \frac{5}{13}, \quad \tan \theta = \frac{5}{12}, \quad \sec \theta = \frac{13}{12} \]