For any angle `theta`, state the value of :
`" "sin^(2)theta+cos^(2)theta`.

Answer

To find the value of \( \sin^2 \theta + \cos^2 \theta \) for any angle \( \theta \), we can follow these steps:

### Step 1: Understand the definitions of sine and cosine
In a right triangle, for an angle \( \theta \):
- The sine of \( \theta \) is defined as the ratio of the length of the opposite side (perpendicular) to the length of the hypotenuse.
- The cosine of \( \theta \) is defined as the ratio of the length of the adjacent side (base) to the length of the hypotenuse.

### Step 2: Write the formulas for sine and cosine
Let:
- \( P \) = length of the perpendicular side
- \( B \) = length of the base side
- \( H \) = length of the hypotenuse

Then:
\[ \sin \theta = \frac{P}{H} \]
\[ \cos \theta = \frac{B}{H} \]

### Step 3: Square both sine and cosine
Now, we square both sine and cosine:
\[ \sin^2 \theta = \left(\frac{P}{H}\right)^2 = \frac{P^2}{H^2} \]
\[ \cos^2 \theta = \left(\frac{B}{H}\right)^2 = \frac{B^2}{H^2} \]

### Step 4: Add the squared values
Now, we add \( \sin^2 \theta \) and \( \cos^2 \theta \):
\[ \sin^2 \theta + \cos^2 \theta = \frac{P^2}{H^2} + \frac{B^2}{H^2} \]

### Step 5: Combine the fractions
Since both terms have the same denominator, we can combine them:
\[ \sin^2 \theta + \cos^2 \theta = \frac{P^2 + B^2}{H^2} \]

### Step 6: Apply the Pythagorean theorem
According to the Pythagorean theorem, we know that:
\[ P^2 + B^2 = H^2 \]
Substituting this into our equation gives:
\[ \sin^2 \theta + \cos^2 \theta = \frac{H^2}{H^2} \]

### Step 7: Simplify the expression
Now, simplifying the fraction:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]

### Final Answer
Thus, for any angle \( \theta \):
\[ \sin^2 \theta + \cos^2 \theta = 1 \]

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