TITLE: Teacher claims this proof for $\frac{\csc\theta-1}{\cot\theta}=\frac{\cot\theta}{\csc\theta+1}$ is wrong. Why?
QUESTION [36 upvotes]: My son's high school teacher says his solution to this proof is wrong because it is not "the right way" and that you have to "start with one side of the equation and prove it is equal to the other". After reviewing it, I disagree. I believe his solution is correct, even if not "the right way", whatever that means. I asked my son how he did it: he cross-multiplied the given identity, simplified it to a known/obvious equality, and then reversed the steps for the proof. This was a graded exam, and the teacher gave him a zero for this problem.
What do you think about my son's solution? Thanks!
Problem: prove the following trigonometric identity
\begin{align*}
\frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{\cot(\theta)}{\csc(\theta)+1}\ .\\
\end{align*}
Solution:  for all real $\theta$ not equal to an integer multiple of $\pi/2$, we have
\begin{align*}
\cot^2(\theta)&=\cot^2(\theta)\\[8pt]
\frac{\cos^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt]
\frac{1-\sin^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt]
\csc^2(\theta)-1&=\cot^2(\theta)\\[8pt]
\frac{\csc^2(\theta)-1}{\cot(\theta)}&=\cot(\theta)\\[8pt]
\frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{\cot(\theta)}{\csc(\theta)+1}
\end{align*}

REPLY [12 votes]: There are multiple motivations for writing proofs: perhaps to pass an exam (from a regular student's viewpoint), or perhaps to test understanding (from a teacher's perspective). Here, I'd like to highlight another important motivation: to enjoy the process of discovery.
Here's a geometric proof of the trigonometric identity. It is not practical to use in an exam, but it makes the abstract symbols more concrete. Coming up with multiple proofs of the same result can help one develop intuition and allow one to appreciate the connections between different areas of mathematics.

Problem statement: $$\frac{\csc\theta-1}{\cot\theta}=\frac{\cot\theta}{\csc\theta+1}$$
In this picture involving the unit circle, we construct some lengths which correspond to expressions in the problem statement. We start with $\theta$ and $\phi = \frac{\pi}{2} - \theta$.
Basic trigonometric definitions lead to the blue and orange lengths. By some angle-chasing (or a circle theorem), we have the $\color{red}{\text{red}}$ angle. Then, more angle-chasing gives:
The $\color{red}{\text{red}}$ angles are the same $\implies\triangle ADB\sim \triangle BDE \implies \frac{AD}{DB}=\frac{DB}{ED}\iff$ the problem statement.