`cos^(2)thetadiv(cot^(2)theta-cos^(2)theta)=3`, Find `tantheta+cosectheta`

Options

A. 1.)`3√5`
B. 2.)`5//√3`
C. 3.) `5√3`
D. 4.) `√3//5`

Answer

To solve the equation \(\frac{\cos^2 \theta}{\cot^2 \theta - \cos^2 \theta} = 3\) and find \( \tan \theta + \csc \theta \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{\cos^2 \theta}{\cot^2 \theta - \cos^2 \theta} = 3 \]
We can rewrite \(\cot^2 \theta\) in terms of sine and cosine:
\[ \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \]
Thus, we can substitute this into our equation:
\[ \frac{\cos^2 \theta}{\frac{\cos^2 \theta}{\sin^2 \theta} - \cos^2 \theta} = 3 \]

### Step 2: Simplify the denominator
Now, simplify the denominator:
\[ \frac{\cos^2 \theta}{\frac{\cos^2 \theta - \cos^2 \theta \sin^2 \theta}{\sin^2 \theta}} = 3 \]
This simplifies to:
\[ \frac{\cos^2 \theta \sin^2 \theta}{\cos^2 \theta (1 - \sin^2 \theta)} = 3 \]
Since \(1 - \sin^2 \theta = \cos^2 \theta\), we have:
\[ \frac{\sin^2 \theta}{\cos^2 \theta} = 3 \]

### Step 3: Relate to tangent
The left side can be rewritten as:
\[ \tan^2 \theta = 3 \]
Taking the square root gives:
\[ \tan \theta = \sqrt{3} \]

### Step 4: Find \(\csc \theta\)
Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we know:
\[ \sin \theta = \sqrt{3} \cos \theta \]
Using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting \(\sin \theta\):
\[ (\sqrt{3} \cos \theta)^2 + \cos^2 \theta = 1 \]
\[ 3 \cos^2 \theta + \cos^2 \theta = 1 \]
\[ 4 \cos^2 \theta = 1 \implies \cos^2 \theta = \frac{1}{4} \implies \cos \theta = \frac{1}{2} \]
Thus, \(\sin \theta = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}\).

### Step 5: Calculate \(\csc \theta\)
Now, we can find \(\csc \theta\):
\[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \]

### Step 6: Find \( \tan \theta + \csc \theta \)
Now, we can find \( \tan \theta + \csc \theta \):
\[ \tan \theta + \csc \theta = \sqrt{3} + \frac{2}{\sqrt{3}} \]
To combine these, we can write \(\sqrt{3}\) as \(\frac{3}{\sqrt{3}}\):
\[ \tan \theta + \csc \theta = \frac{3}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{5}{\sqrt{3}} \]

### Final Answer
Thus, the final answer is:
\[ \tan \theta + \csc \theta = \frac{5}{\sqrt{3}} \]