trignometry

sinA/sinB=p

sin²A/sin²B = p²

sin²A/(1 - cos²B) = p²

But cos²B = cos²A/q²

sin²A/(1 - cos²A/q²) = p²

q²sin²A/(q² - cos²A) = p²

q²sin²A = p²(q² - cos²A) = p²q² - p² + p²sin²A

sin²A = (p²q² - p²)/(q²-p²)

and hence cos²A = (q² - p²q²)/(q²-p²)

tan²A = (p²q² - p²)/(q² - p²q²)

tan A = (±p/q)(q²-1)^1/2/(1-p²)^1/2

and we can see that tanA/tanB = p/q, hence

tan B = ±(q²-1)^1/2/(1-p²)^1/2

Regards,

Team,

TopperLearning.