For the following two (02) items: Let $p\sin^2 \alpha+q\cos^2 \alpha=m$, $q\sin^2 \beta+p\cos^2\beta=n$; $p \neq m, n$ and $q \neq m, n$.
We are given two trigonometric equations:
We are also told that the angles \(\alpha\) and \(\beta\) are complementary. This means their sum is \(90^\circ\) (or \(\frac{\pi}{2}\) radians).
Mathematically, \(\alpha + \beta = 90^\circ\).
From the complementary angle relationship, we know:
Squaring both sides of these identities, we get:
Now, let's substitute these results into the second given equation (\(n\)):
Original Equation 2: \(q\sin^2 \beta+p\cos^2\beta=n\)
Substitute \(\sin^2 \beta = \cos^2 \alpha\) and \(\cos^2 \beta = \sin^2 \alpha\):
\(q(\cos^2 \alpha) + p(\sin^2 \alpha) = n\)
Rearranging the terms to match the format of Equation 1:
\(p\sin^2 \alpha + q\cos^2 \alpha = n\)
Now we have two expressions:
Since the left-hand sides of both expressions are identical (\(p\sin^2 \alpha+q\cos^2 \alpha\)), their right-hand sides must also be equal.
Therefore, we can conclude that \(m = n\).
If \(m = n\), we can rearrange this equation to find the correct relationship among the options.
Subtracting \(n\) from both sides gives:
\(m - n = n - n\)
\(m - n = 0\)
This result matches the fourth option provided.
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