If $\ sin \theta $ and $\ cos \theta$ are roots of the equation $ax^2 +bx +c=0$ then
- (a)$(a-c)^2 = b^2 -c^2$
- (b)$(a-c)^2 = b^2 +c^2$
- (c)$(a+c)^2 = b^2 -c^2$
- (d)$(a+c)^2 = b^2 + c^2$
Roots sin(theta) and cos(theta) imply sin(theta) + cos(theta) = -b/a and sin(theta) * cos(theta) = c/a. Squaring the sum: sin^2 + cos^2 + 2sin*cos = b^2/a^2. 1 + 2(c/a) = b^2/a^2. 1 + 2c/a = b^2/a^2. Multiply by a^2: a^2 + 2ac = b^2. Adding c^2 to both sides: a^2 + 2ac + c^2 = b^2 + c^2, which is (a+c)^2 = b^2 + c^2.
By Vieta's formulas for ax^2 + bx + c = 0, the sum of the roots is -b/a and the product is c/a. Substituting the roots sin(theta) and cos(theta) gives sin(theta) + cos(theta) = -b/a and sin(theta)cos(theta) = c/a. Squaring the sum yields 1 + 2sin(theta)cos(theta), which translates to b^2/a^2 = 1 + 2c/a. Multiplying by a^2 and rearranging terms results in b^2 = a^2 + 2ac, or b^2 = (a + c)^2 - c^2. Therefore, the correct identity is (a + c)^2 = b^2 + c^2.