If `theta` is an acute angle and `sintheta=costheta`, then find the value of `2tan^(2)theta+sin^(2)theta-1`.

Options

A. `1(1)/(2)`
B. `-7`
C. 1
D. 3

Answer

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

1. **Given**: \( \sin \theta = \cos \theta \)

Since \( \sin \theta = \cos \theta \), we can use the identity that relates sine and cosine:

2. **Using the identity**:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = 1 \]

This means that:

3. **Finding the angle**:
\[ \theta = 45^\circ \]

Now we need to calculate the expression \( 2\tan^2 \theta + \sin^2 \theta - 1 \):

4. **Calculate \( \tan^2 \theta \)**:
\[ \tan^2 45^\circ = 1^2 = 1 \]

5. **Calculate \( \sin^2 \theta \)**:
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad \sin^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

6. **Substituting values into the expression**:
\[ 2\tan^2 \theta + \sin^2 \theta - 1 = 2(1) + \frac{1}{2} - 1 \]

7. **Simplifying the expression**:
\[ = 2 + \frac{1}{2} - 1 = 1 + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2} \]

Thus, the final value of the expression \( 2\tan^2 \theta + \sin^2 \theta - 1 \) is:

\[ \frac{3}{2} \]