asin^2x+bcos^2x=c then tan^2x
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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)}
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)}
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Answer:
(c-b)/(a-c)
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