If `sec(5theta-50^(@))=cosec(theta+32^(@))`, then the value of `theta` is: `(0^(@)ltthetalt90^(@))`

Options

A. `33(1^(@))/(3)`
B. `18^(@)`
C. `3(1^(@))/(3)`
D. `30^(@)`

Answer

To solve the equation \( \sec(5\theta - 50^\circ) = \csc(\theta + 32^\circ) \), we can follow these steps:

### Step 1: Use the identity for secant and cosecant
Recall that \( \sec(x) = \csc(90^\circ - x) \). Therefore, we can rewrite the left-hand side:
\[ \sec(5\theta - 50^\circ) = \csc(90^\circ - (5\theta - 50^\circ)) = \csc(90^\circ - 5\theta + 50^\circ) = \csc(140^\circ - 5\theta) \]
Thus, we can rewrite the equation as:
\[ \csc(140^\circ - 5\theta) = \csc(\theta + 32^\circ) \]

### Step 2: Set the angles equal to each other
Since the cosecant function is equal when the angles are equal or differ by \(180^\circ\), we can set up the following equations:
1. \( 140^\circ - 5\theta = \theta + 32^\circ \)
2. \( 140^\circ - 5\theta = 180^\circ - (\theta + 32^\circ) \)

### Step 3: Solve the first equation
Let's solve the first equation:
\[ 140^\circ - 5\theta = \theta + 32^\circ \]
Rearranging gives:
\[ 140^\circ - 32^\circ = \theta + 5\theta \]
\[ 108^\circ = 6\theta \]
\[ \theta = \frac{108^\circ}{6} = 18^\circ \]

### Step 4: Check the second equation
Now let's check the second equation:
\[ 140^\circ - 5\theta = 180^\circ - (\theta + 32^\circ) \]
This simplifies to:
\[ 140^\circ - 5\theta = 180^\circ - \theta - 32^\circ \]
\[ 140^\circ - 5\theta = 148^\circ - \theta \]
Rearranging gives:
\[ 140^\circ - 148^\circ = -\theta + 5\theta \]
\[ -8^\circ = 4\theta \]
\[ \theta = -2^\circ \]
Since \(-2^\circ\) is not in the range \(0^\circ < \theta < 90^\circ\), we discard this solution.

### Conclusion
The only valid solution is:
\[ \theta = 18^\circ \]