Maths c3 harmonic identities question

Can you explain what you're asking it's not very clear what you mean.

If you pause at 3:14, it says -6=rcos alpha, and -10=rsin alpha. My question is why can't you just divide the two like this and find tan alpha, which would give 10/6. Instead, the actual answer is 6/10. I get you must use a particular identity depending on the way the equation is written, but I don't understand why we MUST use that identity. What's wrong with the identity used at the beginning of the video that caused him use the other harmonic identity.

There's nothing wrong with it and him saying that it "isn't working" is extremely misleading. He seems to require a right-angled-triangle to do the question for some reason. Personally I would ignore him.

Try the question yourself using the first identity and see if it works out for you. If you're having trouble then go back to his video and maybe it will make sense?

So is tan alpha = 10/6 right as well or am I just doing something wrong?

Yes that's fine but you'd have to make sure that your solution to $\tan \alpha = \frac{10}{6}$ gives you a negative sine/cosine (using CAST/graphs etc.). If this part confuses you then maybe you should go with his plan.

In exams they always give you the identity to use and the range for $\alpha$ so you probably don't need to worry.

Yeah tbh that part is confusing because sin alpha and cos alpha will be positive when alpha is between 0 and 90, so I can see there must be something weird that happens if you use another identity. I think i'll just stick with his method.

It's not really weird. A Level trig doesn't just stay within 0-90 and I'm sure you're aware that there is more than one solution to $\tan \alpha = \frac{10}{6}$.

I think you're just too used to $0<\alpha<90$ from textbook questions etc. so anything else is confusing.