The value of `theta` , which satisfies the equation `tan^(2)theta+3=3sectheta,0^(@)lethetalt90^(@)` is

Options

A. `15^(@)or0^(@)`
B. `30^(@)or0^(@)`
C. `45^(@)or0^(@)`
D. `60^(@)or0^(@)`

Answer

To solve the equation \( \tan^2 \theta + 3 = 3 \sec \theta \) for \( 0^\circ < \theta < 90^\circ \), we will follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Using this identity, we can express \( \tan^2 \theta \) in terms of \( \sec^2 \theta \):
\[ \tan^2 \theta = \sec^2 \theta - 1 \]
Substituting this into the original equation gives us:
\[ (\sec^2 \theta - 1) + 3 = 3 \sec \theta \]
This simplifies to:
\[ \sec^2 \theta + 2 = 3 \sec \theta \]

### Step 2: Rearranging the equation
Rearranging the equation, we get:
\[ \sec^2 \theta - 3 \sec \theta + 2 = 0 \]

### Step 3: Let \( T = \sec \theta \)
Now, we can substitute \( T \) for \( \sec \theta \):
\[ T^2 - 3T + 2 = 0 \]

### Step 4: Factor the quadratic equation
We can factor the quadratic equation:
\[ (T - 1)(T - 2) = 0 \]
This gives us two possible solutions:
\[ T - 1 = 0 \quad \Rightarrow \quad T = 1 \]
\[ T - 2 = 0 \quad \Rightarrow \quad T = 2 \]

### Step 5: Solve for \( \theta \)
Now, we substitute back for \( \sec \theta \):
1. For \( T = 1 \):
\[ \sec \theta = 1 \quad \Rightarrow \quad \cos \theta = 1 \quad \Rightarrow \quad \theta = 0^\circ \]
2. For \( T = 2 \):
\[ \sec \theta = 2 \quad \Rightarrow \quad \cos \theta = \frac{1}{2} \quad \Rightarrow \quad \theta = 60^\circ \]

### Step 6: Conclusion
Thus, the values of \( \theta \) that satisfy the equation in the interval \( 0^\circ < \theta < 90^\circ \) are:
\[ \theta = 60^\circ \]