Sec(2x)=sec^2x/2-sec^2x prove the identity pleaseeeeeeeeeeeee?
sec(2x) = sec²x / (2-sec²x) let us express it in terms of sinx, cosx: [1/cos(2x)] = (1/cos²x) / [2 - (1/cos²x)] → you know, according to double-angle identities, that: cos(2x) = cos²x - sin²x thus: 1/ (cos²x - sin²x) = (1/cos²x) / [2 - (1/cos²x)] → 1/ (co...
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