Question

if sinA and cosA are roots of ax²+bx+c=0 then, prove a²+2ac=b²​

Answer 1

Given :

• sin A and cos A are roots of equation ax² + bx + c = 0

To prove :

• a² + 2 ac = b²

Knowledge required :

For a quadratic equation of the form ax² + bx + c = 0

$\red{\bullet}\;\sf{sum\:of\:zeroes=\dfrac{-(coefficient\:of\;x)}{coefficient\;of\:x^2}=\dfrac{-b}{a}}$

$\red{\bullet}\;\sf{product\:of\:zeroes=\dfrac{constant\:term}{coefficient\;of\:x^2}=\dfrac{c}{a}}$

Solution :

Since, sin A and cos A are zeroes of ax² + bx + c = 0

so,

$\red{\longrightarrow}\sf{sin\;A+cos\;A=\dfrac{-b}{a}\;\;\;\;\;equation (1)}$

and,

$\red{\longrightarrow}\sf{sin\;A\;cos\;A=\dfrac{c}{a}\;\;\;\;\;\;\;equation(2)}$

Now,

Using algebraic identity

( m + n )² = m² + n²  + 2 m n

$\red{\longrightarrow}\sf{(sin\;A+cos\;A)^2=sin^2A+cos^2A+2\;sin\;A\;cos\;A}$

using equation (1) and equation (2)

$\red{\longrightarrow}\sf{\left(\dfrac{-b}{a}\right)^2=sin^2A+cos^2A+2\left(\;\dfrac{c}{a}\;\right)}$

using trigonometric identity

sin²Ф + cos²Ф = 1

$\red{\longrightarrow}\sf{\left(\dfrac{-b}{a}\right)^2= (1) +2\left(\;\dfrac{c}{a}\;\right)}$

$\red{\longrightarrow}\sf{\dfrac{b^2}{a^2}= 1 +\dfrac{2c}{a}}$

$\red{\longrightarrow}\sf{\dfrac{b^2}{a^2}=\dfrac{a+2c}{a}}$

$\red{\longrightarrow}\sf{\dfrac{b^2}{a}=a+2c}$

$\overbrace{\underbrace{\boxed{\red{\longrightarrow}\sf{b^2=a^2+2ac}}}}$

PROVED .

Answer 2

♡REFER THIS ATTACHMENT♡