If \(\sec^{2}\alpha = \frac{4ab}{(a+b)^{2}}\), then:
\(a = b\)
For any angle α, sec2α ≥ 1.
So \(\frac{4ab}{(a+b)^{2}} \ge 1\), which gives (a + b)2 ≤ 4ab.
Expanding, a2 + 2ab + b2 ≤ 4ab, so a2 − 2ab + b2 ≤ 0, that is (a − b)2 ≤ 0.
A square can never be negative, so (a − b)2 = 0 and a = b. (Checking: with a = b the fraction is 4a2/4a2 = 1 = sec20°.)
Hence, the answer is \(a = b\).
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