# How is the trig identity $\tan^2(x) = \sec^2(x) -1$ derived? [closed]

I was practicing some integration when I came across a problem that required the trig identity $$\tan^2(x) = \sec^2(x)-1$$. However, I have never come across this identity and would like to know how it's derived.

## closed as off-topic by TheSimpliFire, Cornman, Leucippus, Lee David Chung Lin, Eevee TrainerMay 7 at 6:52

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Divide the identity $$\sin ^2x + \cos^2 x = 1$$ by $$\cos^2 x$$.

$$\frac{\sin ^2x + \cos^2 x }{\cos^2 x}= \frac{1}{\cos^2 x}$$

$$\frac{\sin ^2x}{\cos^2x}+ \frac{\cos^2 x }{\cos^2 x}= \frac{1}{\cos^2 x}$$

$$\tan ^2x + 1 = \sec^2 x$$

$$\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{\sin^2 x + \cos^2 x - \cos^2x}{\cos^2x} = \frac{1-\cos^2x}{\cos^2x} = \frac{1}{\cos^2x} - 1 = \sec^2x -1$$

By Pythagoras' theorem, we have $$(AB)^2+(BC)^2=(AC)^2$$

Divide the above equation by $$(AB)^2$$ to see $$\frac{(AB)^2}{(AB)^2}+\frac{(BC)^2}{(AB)^2}=\frac{(AC)^2}{(AB)^2} \Longrightarrow1+\Bigg(\frac{BC}{AB}\Bigg)^2=\Bigg(\frac{AC}{AB}\Bigg)^2$$ so $$1+\tan ^2 \theta=\sec^2 \theta$$ where $$\theta=\angle CAB$$