The image shows the following problem:

5. If $\sin \theta = \cos \theta$, find the value of $2 \tan^2 \theta + 3 \sin^2 \theta - 1$.

Let's solve this step-by-step.

Step 1: Solving $\sin \theta = \cos \theta$

We know that $\sin \theta = \cos \theta$ implies:

$\frac{\sin \theta}{\cos \theta} = 1$

This simplifies to:

$\tan \theta = 1$

Thus, $\theta = 45^\circ$ or $\theta = \frac{\pi}{4}$ radians.

Step 2: Calculate $2 \tan^2 \theta + 3 \sin^2 \theta - 1$

Since $\tan \theta = 1$, we can substitute into the expression:

$\tan^2 \theta = 1^2 = 1$ $\sin \theta = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ $\sin^2 \theta = \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{2}{4} = \frac{1}{2}$

Now, substitute these values into the expression:

$2 \tan^2 \theta + 3 \sin^2 \theta - 1 = 2(1) + 3\left( \frac{1}{2} \right) - 1$ $= 2 + \frac{3}{2} - 1$ $= 2 - 1 + \frac{3}{2} = 1 + \frac{3}{2} = \frac{2}{2} + \frac{3}{2} = \frac{5}{2}$

Final Answer:

The value of the expression is $\frac{5}{2}$.

Would you like any more details or explanations? Here are some related questions you might consider:

1. What is the general solution to $\sin \theta = \cos \theta$?
2. How does the identity $\tan^2 \theta + 1 = \sec^2 \theta$ relate to this problem?
3. What are the exact values of trigonometric functions for other angles like $30^\circ$ or $60^\circ$?
4. How do you convert between radians and degrees?
5. What is the double-angle formula for sine, and how is it derived?

Tip: Always remember that $\sin \theta = \cos \theta$ occurs when $\theta = 45^\circ + n \cdot 180^\circ$ where $n$ is an integer, considering the periodicity of trigonometric functions.