If `sintheta+cosec theta=2`, then `sin^(2)theta+cosec ^(2)theta` is equal to

Options

A. `1`
B. `4`
C. `2`
D. None of these

Explanation

To solve the problem, we start with the given equation:

**Given:**
\[ \sin \theta + \csc \theta = 2 \]

**Step 1: Rewrite \(\csc \theta\) in terms of \(\sin \theta\)**
\[ \csc \theta = \frac{1}{\sin \theta} \]
So, we can rewrite the equation as:
\[ \sin \theta + \frac{1}{\sin \theta} = 2 \]

**Step 2: Multiply through by \(\sin \theta\) to eliminate the fraction**
\[ \sin^2 \theta + 1 = 2\sin \theta \]

**Step 3: Rearrange the equation**
\[ \sin^2 \theta - 2\sin \theta + 1 = 0 \]

**Step 4: Recognize this as a perfect square**
The left-hand side can be factored:
\[ (\sin \theta - 1)^2 = 0 \]

**Step 5: Solve for \(\sin \theta\)**
Taking the square root of both sides gives:
\[ \sin \theta - 1 = 0 \implies \sin \theta = 1 \]

**Step 6: Find \(\csc \theta\)**
Since \(\sin \theta = 1\), we have:
\[ \csc \theta = \frac{1}{\sin \theta} = 1 \]

**Step 7: Calculate \(\sin^2 \theta + \csc^2 \theta\)**
Now we need to find:
\[ \sin^2 \theta + \csc^2 \theta \]
Substituting the values we found:
\[ \sin^2 \theta = 1^2 = 1 \]
\[ \csc^2 \theta = 1^2 = 1 \]
Thus:
\[ \sin^2 \theta + \csc^2 \theta = 1 + 1 = 2 \]

**Final Answer:**
\[ \sin^2 \theta + \csc^2 \theta = 2 \]

---