If `(cos^(2)theta)/(cot^(2)theta-cos^(2)theta)=3and0^(@)ltthetalt90^(@)`, then find the value of `theta`.

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. None of these

Answer

To solve the equation \(\frac{\cos^2 \theta}{\cot^2 \theta - \cos^2 \theta} = 3\) where \(0^\circ < \theta < 90^\circ\), we will follow these steps:

### Step 1: Rewrite cotangent in terms of sine and cosine
Recall that \(\cot \theta = \frac{\cos \theta}{\sin \theta}\). Therefore, we can express \(\cot^2 \theta\) as:
\[ \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \]

### Step 2: Substitute cotangent into the equation
Substituting \(\cot^2 \theta\) into the equation gives:
\[ \frac{\cos^2 \theta}{\frac{\cos^2 \theta}{\sin^2 \theta} - \cos^2 \theta} = 3 \]

### Step 3: Simplify the denominator
The denominator can be simplified:
\[ \frac{\cos^2 \theta}{\frac{\cos^2 \theta - \cos^2 \theta \sin^2 \theta}{\sin^2 \theta}} = 3 \]
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:
\[ \frac{\cos^2 \theta \sin^2 \theta}{\cos^2 \theta \cos^2 \theta} = 3 \]

### Step 4: Cancel out \(\cos^2 \theta\)
Assuming \(\cos^2 \theta \neq 0\) (which is valid since \(0^\circ < \theta < 90^\circ\)), we can cancel \(\cos^2 \theta\) from the numerator and denominator:
\[ \frac{\sin^2 \theta}{\cos^2 \theta} = 3 \]

### Step 5: Rewrite in terms of tangent
This can be rewritten as:
\[ \tan^2 \theta = 3 \]

### Step 6: Solve for \(\tan \theta\)
Taking the square root of both sides, we get:
\[ \tan \theta = \sqrt{3} \]
Since \(\theta\) is in the range \(0^\circ < \theta < 90^\circ\), we take the positive root.

### Step 7: Find the angle \(\theta\)
The angle whose tangent is \(\sqrt{3}\) is:
\[ \theta = 60^\circ \]

### Final Answer
Thus, the value of \(\theta\) is:
\[ \theta = 60^\circ \]

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