If `cosec^(2)theta+cot^(2)theta=7` find `sintheta&costheta`

Answer

To solve the equation \( \csc^2 \theta + \cot^2 \theta = 7 \) and find \( \sin \theta \) and \( \cos \theta \), we can follow these steps:

### Step 1: Use the identity for \( \csc^2 \theta \) and \( \cot^2 \theta \)
We know the identity:
\[ \csc^2 \theta - \cot^2 \theta = 1 \]

### Step 2: Set up the equations
From the given equation:
\[ \csc^2 \theta + \cot^2 \theta = 7 \quad \text{(1)} \]
And from the identity:
\[ \csc^2 \theta - \cot^2 \theta = 1 \quad \text{(2)} \]

### Step 3: Add equations (1) and (2)
Adding both equations:
\[ (\csc^2 \theta + \cot^2 \theta) + (\csc^2 \theta - \cot^2 \theta) = 7 + 1 \]
This simplifies to:
\[ 2 \csc^2 \theta = 8 \]

### Step 4: Solve for \( \csc^2 \theta \)
Dividing both sides by 2:
\[ \csc^2 \theta = 4 \]

### Step 5: Find \( \sin^2 \theta \)
Since \( \csc \theta = \frac{1}{\sin \theta} \), we have:
\[ \csc^2 \theta = \frac{1}{\sin^2 \theta} \]
Thus,
\[ \frac{1}{\sin^2 \theta} = 4 \]
Taking the reciprocal gives:
\[ \sin^2 \theta = \frac{1}{4} \]

### Step 6: Find \( \sin \theta \)
Taking the square root:
\[ \sin \theta = \frac{1}{2} \quad \text{(considering the principal value)} \]

### Step 7: Find \( \cos^2 \theta \)
Using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting \( \sin^2 \theta \):
\[ \frac{1}{4} + \cos^2 \theta = 1 \]
Thus,
\[ \cos^2 \theta = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 8: Find \( \cos \theta \)
Taking the square root:
\[ \cos \theta = \frac{\sqrt{3}}{2} \quad \text{(considering the principal value)} \]

### Final Answer
Thus, we have:
\[ \sin \theta = \frac{1}{2} \quad \text{and} \quad \cos \theta = \frac{\sqrt{3}}{2} \]

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