Verification of proof of $\tan(x)+\cot(x)=\csc(x)\sec(x)$

Question

So I used to proof tan(x)+cot(x)=csc(x)sec(x) by derivation from pythagorean identity? so Is my proof Right?

Answer 1

Seems fine, we can combine the division togeher.

From $$\sin^2 x + \cos^2 x = 1,$$

divide by $\sin x \cos x$,

$$\frac{\sin^2 x}{\sin x \cos x} + \frac{\cos^2 x}{\sin x \cos x} = \frac1{\sin x \cos x},$$

$$\tan x + \cot x = \sec x \operatorname{cosec} x$$

Answer 2

Looks fine to me. The division by $\sin x$ and $\cos x$ could fail (either of those could be 0), but that only happens where the original expression is undefined anyway.

Answer 3

Your argument is quite confused, but contains the things that ought to appear.

You can streamline it: \begin{align} \tan x+\cot x &=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} \\ &=\frac{\sin^2x+\cos^2x}{\cos x\sin x} \\ &=\frac{1}{\cos x\sin x} \\ &=\sec x\csc x \end{align}