Question

If sin theta & cos theta are roots of the equations ax2 + bx+c=0
prove that a ²-b2 +2ac=0.​

Answer 1

Answer:

Given:

$\sf{sin\theta \ and \ cos\theta \ are \ roots \ of}$

$\sf{the \ equation \ ax^{2}+bx+c=0}$

Solution:

$\sf{Sum \ of \ roots=\dfrac{b}{a}}$

$\sf{\therefore{sin\theta+cos\theta=\dfrac{b}{a}...(1)}}$

$\sf{Product \ of \ roots=\dfrac{c}{a}}$

$\sf{\therefore{sin\theta.cos\theta=\dfrac{c}{a}...(2)}}$

$\sf{From \ equation (1) \ we \ get,}$

$\sf{\dfrac{b}{a}=sin\theta+cos\theta}$

$\sf{\therefore{b=a(sin\theta+cos\theta)}}$

$\sf{On \ squaring \ both \ sides \ we \ get,}$

$\sf{\longmapsto{b^{2}=a^{2}(sin\theta+cos\theta)^{2}}}$

$\sf{\underline{\underline{To \ prove:}}}$

$\sf{a^{2}-b^{2}+2ac=0}$

$\sf{\underline{\underline{Proof:}}}$

$\sf{L.H.S.=a^{2}-b^{2}+2ac}$

$\sf{Substitute \ b^{2}=a^{2}(sin\theta+cos\theta)^{2}}$

$\sf{=a^{2}-a^{2}(sin\theta+cos\theta)^{2}+2ac}$

$\sf{By \ identity \ (a+b)^{2}=a^{2}+b^{2}+2ab}$

$\sf{=a^{2}-a^{2}(sin^{2}\theta+cos^{2}\theta+2sin\theta.cos\theta)+2ac}$

$\sf{By \ Trigonometric \ identity}$

$\sf{sin^{2}\theta+cos^{2}\theta=1}$

$\sf{=a^{2}-a^{2}(1+2sin\theta.cos\theta)+2ac}$

$\sf{From \ equation (3), \ we \ get}$

$\sf{=a^{2}-a^{2}(1+\dfrac{2c}{a})+2ac}$

$\sf{=a^{2}-a^{2}-2ac+2ac}$

$\sf{=0}$

$\sf{=R.H.S.}$

$\sf{\underline{\underline{Hence, \ proved.}}}$

$\sf\purple{\tt{a^{2}-b^{2}+2ac=0}}$