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Show that the equation sec^2 theta=(4xy)...

Show that the equation `sec^2 theta=(4xy)/(x+y)^2` is only possible when x=y

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`sec^2theta = (4xy)/(x+y)^2`
We know, `-1 le cos theta le 1`
`=> 0 le cos^2 theta le 1`
`=> sec^2 theta ge 1`.
` (4xy)/(x+y)^2 ge 1`
`=> (4xy)/(x^2+y^2+2xy) ge 1`
`=> (4xy)/(x^2+y^2-2xy + 4xy) ge 1`
`=> (4xy)/((x-y)^2 + 4xy) ge 1`
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