Proving $1+\tan^2(x)=\sec^2(x)$

Question

If one is asked to prove $1+\tan^2(x)=\sec^2(x)$, this is how I would prove it. $$\frac{d}{dx} \frac{\sin(x)}{\cos(x)}=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}=\sec^2(x)$$ and $$\frac{d}{dx} \frac{\sin(x)}{\cos(x)}=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=1+\frac{\sin^2(x)}{\cos^2(x)}=1+\left(\frac{\sin(x)}{\cos(x)}\right)^2=1+\tan^2(x)$$ Therefore $1+\tan^2(x)=\sec^2(x)$ So is my prove valid, and are there any easier ways to prove this?

Answer 1

Note simply that

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