If `4-2sin^(2)theta-5costheta=0,0^(@)ltthetalt90^(@)`, then the value of `(sintheta+tantheta)` is :

Options

A. `(3sqrt(2))/(2)`
B. `(3sqrt(3))/(2)`
C. `3sqrt(2)`
D. `2sqrt(3)`

Answer

To solve the equation \( 4 - 2\sin^2\theta - 5\cos\theta = 0 \) for \( 0^\circ < \theta < 90^\circ \) and find the value of \( \sin\theta + \tan\theta \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 4 - 2\sin^2\theta - 5\cos\theta = 0 \]
Using the identity \( \sin^2\theta = 1 - \cos^2\theta \), we can rewrite the equation in terms of \( \cos\theta \):
\[ 4 - 2(1 - \cos^2\theta) - 5\cos\theta = 0 \]

### Step 2: Simplify the equation
Expanding the equation gives:
\[ 4 - 2 + 2\cos^2\theta - 5\cos\theta = 0 \]
This simplifies to:
\[ 2 + 2\cos^2\theta - 5\cos\theta = 0 \]
Rearranging it, we have:
\[ 2\cos^2\theta - 5\cos\theta + 2 = 0 \]

### Step 3: Solve the quadratic equation
Let \( x = \cos\theta \). The equation becomes:
\[ 2x^2 - 5x + 2 = 0 \]
We can solve this quadratic equation using the factorization method. We need two numbers that multiply to \( 2 \times 2 = 4 \) and add to \( -5 \). The numbers are \( -4 \) and \( -1 \):
\[ 2x^2 - 4x - x + 2 = 0 \]
Factoring by grouping:
\[ 2x(x - 2) - 1(x - 2) = 0 \]
This gives:
\[ (2x - 1)(x - 2) = 0 \]

### Step 4: Find the values of \( x \)
Setting each factor to zero gives:
1. \( 2x - 1 = 0 \) → \( x = \frac{1}{2} \)
2. \( x - 2 = 0 \) → \( x = 2 \)

Since \( \cos\theta \) cannot be greater than 1, we discard \( x = 2 \). Thus:
\[ \cos\theta = \frac{1}{2} \]

### Step 5: Find the angle \( \theta \)
The angle \( \theta \) that satisfies \( \cos\theta = \frac{1}{2} \) in the range \( 0^\circ < \theta < 90^\circ \) is:
\[ \theta = 60^\circ \]

### Step 6: Calculate \( \sin\theta + \tan\theta \)
Now, we find \( \sin\theta \) and \( \tan\theta \):
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]
\[ \tan 60^\circ = \sqrt{3} \]
Thus:
\[ \sin\theta + \tan\theta = \frac{\sqrt{3}}{2} + \sqrt{3} \]
To combine these, we convert \( \sqrt{3} \) to a fraction:
\[ \sqrt{3} = \frac{2\sqrt{3}}{2} \]
So:
\[ \sin\theta + \tan\theta = \frac{\sqrt{3}}{2} + \frac{2\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]

### Final Answer
The value of \( \sin\theta + \tan\theta \) is:
\[ \frac{3\sqrt{3}}{2} \]