Using the trigonometric identity (sinx)^2 + (cosx)^2 = 1, show that (secx)^2 = (tanx)^2 + 1 is also a trigonometric identity.
We can divide by (cosx)^2 across the identity (sinx)^2 + (cosx)^2 = 1 (which can be derived from properties of the unit circle and a bit of Pythagoras’ theorem) to achieve
[(sinx)^2 / (cosx)^2] + [(cosx)^2 / (cosx)^2] = [1 / (cosx)^2]
Which leaves us with our desired identity
(tanx)^2 + 1 (secx)^2 = 1
AB
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