If `cos20^(@)=mandcos70^(@)=n` , then the value of `m^(2)+n^(2)` is

Options

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

Answer

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

1. Let \( m = \cos 20^\circ \)
2. Let \( n = \cos 70^\circ \)

We need to find the value of \( m^2 + n^2 \).

### Step 1: Express \( n \) in terms of sine
Using the co-function identity, we know that:
\[ \cos 70^\circ = \sin 20^\circ \]
Thus, we can express \( n \) as:
\[ n = \sin 20^\circ \]

### Step 2: Substitute \( n \) into the expression
Now we can rewrite \( m^2 + n^2 \):
\[ m^2 + n^2 = \cos^2 20^\circ + \sin^2 20^\circ \]

### Step 3: Apply the Pythagorean identity
According to the Pythagorean identity, we know that:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
Therefore, substituting \( \theta = 20^\circ \):
\[ m^2 + n^2 = \cos^2 20^\circ + \sin^2 20^\circ = 1 \]

### Final Answer
Thus, the value of \( m^2 + n^2 \) is:
\[ \boxed{1} \]

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