Question

If sine, cos are the roots of the equation
ax? - bx + c = 0 then the relation among
a,b,c is
(A) a? - b2 + 2ac = 0
(B) a2 + b2 - 2ac = 0
(C) a? - b2 - 2ac = 0
(D) a2 + b2 + 2ac = 0​

Answer 1

Answer:

A

Step-by-step explanation:

We know

cos²x + sin²x = 1

Here roots are cos and sin

let them be m and n

• m= cosx
• n= sinx

we know m²+n²= 1

From equation sum of roots

• m+n = -(-b/a). = b/a

Product of roots

• m*n = c/a

(m+n)²= m²+n²+2mn

==> m²+n²= (m+n)²- 2mn

==> 1 = (b/a)² - 2(c/a)

==>( b²-2ac ) / a²= 1

==> b²- 2ac= a²

==> a²-b²+2ac = 0

Answer 2

Correct Question:-

If sin θ and cos θ are the roots of the equation ax2 – bx + c = 0, then a, b and c satisfy the relation.

(A) a2 + b2 + 2ac = 0

(B) a2 – b2 + 2ac = 0

(C) a2 + c2 + 2ab = 0

(D) a2 – b2 – 2ac = 0

Answer:-

The correct choice is (B).

Solution:-

Given that

$\bf {sin \: \: \theta \: and \: \: cos \: \: \theta \: are \: \: the \: \: roots \: \: of \: \: \: the \: \: equation \: \: {a}^{2} \: - bx \: + c = 0}$

so

$\bf {sin \: θ + cos \: θ =b/a}$

and

$\bf {sin θ \: cos θ = c/a}$

Using the identity

$\bf \fbox \green{</p><p>(sinθ + cos θ)2 = sin2θ + cos2θ + 2 sin θ cos θ,}$

we have,

$\bf {\fbox \red{ \: \: \: \: \: \: \frac{ {b}^{2} }{ {a}^{2} } = 1 + \frac{2c}{a} \: \: \: \: \: \: } \: \: \: \: \: \: \: \: \: \: or} \: \: \: \: \: \: \: \: \: \: \bf \fbox \red{ \: \: \: \: {a}^{2} - {b}^{2} + 2ac = 0 \: \: \: }$