Question

asin^2x+bcos^2x=c then tan^2x

Answer 1

asin^2x+bcos^2x=c Now Dividing both sides by cos^2x we have
asin^2x/cos^2x+bcos^2x/cos^2x=c/cos^2x
Or,atan^2x+b=csec^2x
Or, atan^2x+b=c(1+tan^2x)(since sec^2x=1+tan^2x)
Or, atan^2x+b=c+ctan^2x
Or, atan^2x-ctan^2x=c-b
Or, (a-c)tan^2x=c-b
Or, tan^2x=(c-b)/(a-c)
Or, tanx=+-√{(c-b)/(a-c)}

Answer 2

Answer:

(c-b)/(a-c)

Step-by-step explanation:

HOPE YOU MARK THE BRAINLIST