<s>[INST] Prove $1 + \tan^2\theta = \sec^2\theta$ Prove the following trigonometric identity: $$1 + \tan^2\theta = \sec^2\theta$$
I'm curious to know of the different ways of proving this depending on different characterizations of tangent and secant.
 [/INST] Assuming the First Pythagorean Trigonometric Identity,
$$\sin^2\theta + \cos^2\theta = 1$$
Dividing by $\cos^2\theta$,
$$\Rightarrow \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$$ 
$$\Rightarrow \left(\frac{\sin\theta}{\cos\theta}\right)^2 + \left(\frac{\cos\theta}{\cos\theta}\right)^2 = \left(\frac{1}{\cos\theta}\right)^2$$ 
Since $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\sec\theta = \dfrac{1}{\cos\theta}$,
$$\Rightarrow \tan^2\theta + 1 = \sec^2\theta $$
Hence Proved.
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