验证$\tan(x)+\cot(x)=\csc(x)\sec(x)$的证明：勾股恒等式推导是否正确？

Hey there! Let’s break this down to confirm if your proof using the Pythagorean identity is valid.

Absolutely—using the Pythagorean identity to prove $\tan{x}+\cot{x}=\csc{x}\sec{x}$ is a totally valid, rigorous approach. Let’s walk through a standard version of this proof (aligning with the logic you’d use) to verify:

Step 1: Rewrite all terms in sine and cosine #

Start by translating each trigonometric function to its basic definition:

• $\tan{x} = \frac{\sin{x}}{\cos{x}}$
• $\cot{x} = \frac{\cos{x}}{\sin{x}}$
• $\csc{x} = \frac{1}{\sin{x}}$
• $\sec{x} = \frac{1}{\cos{x}}$

The left-hand side (LHS) of the equation becomes:
$\tan{x} + \cot{x} = \frac{\sin{x}}{\cos{x}} + \frac{\cos{x}}{\sin{x}}$

Step 2: Combine the fractions #

Find a common denominator ($\sin{x}\cos{x}$) to add the two terms:
$\frac{\sin^2{x} + \cos^2{x}}{\sin{x}\cos{x}}$

Step 3: Apply the Pythagorean identity #

This is where the key identity comes in: $\sin^2{x} + \cos^2{x} = 1$. Substitute this into the numerator:
$\frac{1}{\sin{x}\cos{x}}$

Step 4: Split the fraction to match the right-hand side (RHS) #

Rewrite the result as a product of reciprocals to match the form of $\csc{x}\sec{x}$:
$\frac{1}{\sin{x}} \cdot \frac{1}{\cos{x}} = \csc{x}\sec{x}$

Since we’ve shown LHS = RHS through these logical steps, your proof using the Pythagorean identity is completely correct—as long as you followed a similar flow and didn’t skip critical rigor (like defining domain restrictions, where $x$ can’t be a multiple of $\pi/2$ or $\pi$, if you need full thoroughness).

内容的提问来源于stack exchange，提问作者Moayd Shagaf