Solution:

Given equation is ax² + bx + c = 0

Let tan²A and sin²A be the roots of the equation.

We know that tan²A + sin²A = 1

Also, we know that tanA = √(tan²A) and sinA = √(sin²A)

Hence, tanA + sinA = √(tan²A) + √(sin²A) = √(tan²A + sin²A) = √1 = 1

Now, let's find the sum and product of the roots.

Sum of roots = tan²A + sin²A = 1

Product of roots = tan²A × sin²A = (tanA × sinA)² = (sinA/cosA × sinA)² = sin⁴A/cos²A

We can write cos²A = 1 - sin²A

Therefore, product of roots = sin⁴A/(1 - sin²A)

Now, let's use the Vieta's formula to find b and c.

Sum of roots = -b/a

1 = -b/a

b = -a

Product of roots = c/a

sin⁴A/(1 - sin²A) = c/a

c = a sin⁴A/(1 - sin²A)

Now, let's find b² - c².

b² - c² = (-a)² - [a sin⁴A/(1 - sin²A)]²

= a² - a² sin⁸A/(1 - 2sin²A + sin⁴A)

= a²(1 - sin⁸A/(1 - 2sin²A + sin⁴A))

= a²(1 - sin⁴A/(1 - sin²A))^2

= a²(cos²A)^2

= a²cos⁴A

= (a cos²A)²

= (a - b)²

= (-b)²

= a²

Therefore, b² - c² = a²

Answer: b² - c² = a²

Ax²+bx+c=0 is the given equation
and roots are tan²x, sin²x

sum of roots : tan²x+sin²x = -b/a
( sin²x+sin²xcos²x)/cos²x =-b/a
sin²x(1+cos²x) /cos²x=-b/a
(1-cos²x) (1+cos²x)/cos²x=-b/a
(1-cos4x) /cos2x=-b/a
product of roots :tan2xsin2x=c/a
sin4x/cos2x=c/a
1-cos4x /cos2x=c/a
so,
-b/a=c/a
-b=c
squaring on both sides
b²-c²=0
hope it helps you