IF `2 sin theta-8 cos^2 theta+5=0, 0^@ lt theta lt 90^@` then what is the value of `(tan 2 theta+ cosec 2 theta)`

Options

A. `3 sqrt3`
B. `(4 sqrt3)/3`
C. `(5 sqrt3)/3`
D. `2 sqrt3`

Answer

To solve the equation \(2 \sin \theta - 8 \cos^2 \theta + 5 = 0\) for \(0^\circ < \theta < 90^\circ\) and find the value of \(\tan 2\theta + \csc 2\theta\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2 \sin \theta - 8 \cos^2 \theta + 5 = 0 \]
We know that \(\cos^2 \theta = 1 - \sin^2 \theta\). Thus, we can rewrite the equation in terms of \(\sin \theta\):
\[ 2 \sin \theta - 8(1 - \sin^2 \theta) + 5 = 0 \]
Expanding this gives:
\[ 2 \sin \theta - 8 + 8 \sin^2 \theta + 5 = 0 \]
Simplifying, we have:
\[ 8 \sin^2 \theta + 2 \sin \theta - 3 = 0 \]

### Step 2: Solve the quadratic equation
Now we can solve the quadratic equation \(8 \sin^2 \theta + 2 \sin \theta - 3 = 0\) using the quadratic formula:
\[ \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 8\), \(b = 2\), and \(c = -3\):
\[ \sin \theta = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 8 \cdot (-3)}}{2 \cdot 8} \]
Calculating the discriminant:
\[ \sin \theta = \frac{-2 \pm \sqrt{4 + 96}}{16} = \frac{-2 \pm \sqrt{100}}{16} = \frac{-2 \pm 10}{16} \]
This gives us two possible solutions:
\[ \sin \theta = \frac{8}{16} = \frac{1}{2} \quad \text{or} \quad \sin \theta = \frac{-12}{16} = -\frac{3}{4} \]
Since \(\sin \theta\) must be positive in the interval \(0^\circ < \theta < 90^\circ\), we have:
\[ \sin \theta = \frac{1}{2} \]

### Step 3: Find \(\theta\)
From \(\sin \theta = \frac{1}{2}\), we find:
\[ \theta = 30^\circ \]

### Step 4: Calculate \(\tan 2\theta\) and \(\csc 2\theta\)
Now we need to calculate \(\tan 2\theta\) and \(\csc 2\theta\):
\[ 2\theta = 60^\circ \]
Thus:
\[ \tan 2\theta = \tan 60^\circ = \sqrt{3} \]
And:
\[ \csc 2\theta = \csc 60^\circ = \frac{2}{\sqrt{3}} \]

### Step 5: Find \(\tan 2\theta + \csc 2\theta\)
Now we add these two values:
\[ \tan 2\theta + \csc 2\theta = \sqrt{3} + \frac{2}{\sqrt{3}} \]
To combine these, we can rationalize:
\[ = \sqrt{3} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{5}{\sqrt{3}} \]
Thus, the final answer is:
\[ \tan 2\theta + \csc 2\theta = \frac{5}{\sqrt{3}} \]