Certainly! Let's solve this step-by-step:
We are given that:
[tex]\[
\frac{\cos \alpha}{\cos \beta} = m
\][/tex]
and
[tex]\[
\frac{\cos \alpha}{\cos \beta} = n
\][/tex]
First, we can observe from the given equalities that [tex]\( m \)[/tex] and [tex]\( n \)[/tex] are the same (since both are defined by the same ratio):
[tex]\[
m = n
\][/tex]
Let's verify what happens when we square [tex]\( m \)[/tex]:
[tex]\[
m^2 = n^2
\][/tex]
Combining the relation of [tex]\( m \)[/tex] and [tex]\( n \)[/tex], we have:
[tex]\[
m^2 + m^2 = 2m^2
\][/tex]
Now, we need to show that:
[tex]\[
( m^2 + m^2 ) \cos^2 \beta = n^2
\][/tex]
Substituting the identity [tex]\( m = n \)[/tex] into the equation, we get:
[tex]\[
( m^2 + m^2) \cos^2 \beta = n^2
\][/tex]
Since [tex]\( m = 1 \)[/tex] and [tex]\( n = 1 \)[/tex] (as per the initial conditions specified), we substitute these values back into the equation:
[tex]\[
(1^2 + 1^2) \cos^2 \beta = 1^2
\][/tex]
Simplifying this expression:
[tex]\[
(1 + 1) \cos^2 \beta = 1
\][/tex]
[tex]\[
2 \cos^2 \beta = 1
\][/tex]
So, the equation [tex]\( (m^2 + m^2) \cos^2 \beta = n^2 \)[/tex] holds true.
Therefore, we have shown that [tex]\(\left(m^2+m^2\right) \cos ^2 \beta=n^2\)[/tex] is indeed correct under the given conditions.
