Answer:
Step-by-step explanation:

Hint: 1 - Cos² A = Sin² A
Answer:
your rue
Step-by-step explanation:
Answer:
3
Step-by-step explanation:
Recall that

Dividing both sides by cosh²(x) gives

Also, recall the identity

Then

Answer:
The answer is 8 : 22, 12:33 as they are two equelivant ratios 11 and 4