If sin α and cos α are the roots of the equations* *ax^{2}^{ }+ bx + c = 0, then b^{2} =

#### Options

^{2}− 2aca

^{2}+ 2aca

^{2}− aca

^{2}+ ac

#### Solution

The given quadric equation is `ax^2 + bx + c = 0`, and `sin alpha and cos beta` are roots of given equation.

And, a = a,b = b and, c = c

Then, as we know that sum of the roots

`sin alpha + cos beta - (-b)/a`…. (1)

And the product of the roots

`sin alpha .cos beta =c/a`…. (2)

Squaring both sides of equation (1) we get

`(sin alpha + cos beta)^2 = ((-b)/a)^2`

`sin^2 alpha + cos^2 beta + 2 sin alpha cos beta = b^2/a^2`

Putting the value of `sin^2 alpha + cos^2 beta = 1`, we get

`1 + 2 sin alpha cos beta = b^2/a^2`

`a^2 (1+2 sin alpha cos beta) = b^2`

Putting the value of`sin alpha.cos beta = c/a` , we get

`a^2 (1 + 2 c/a) = b^2`

`a^2 ((a+2c)/a) = b^2`

`a^2 + 2ac =b^2`

Therefore, the value of `b^2 = a^2 + 2ac`.