Question

If \[\sin \alpha \] and \[\cos \alpha \] are the roots of the equation \[l{{x}^{2}}+mx+n=0\], then:
A. \[{{l}^{2}}-{{m}^{2}}+2lm=0\]
B. \[{{l}^{2}}+{{m}^{2}}+2\ln =0\]
C. \[{{l}^{2}}-{{m}^{2}}+2\ln =0\]
D. \[{{l}^{2}}+{{m}^{2}}-2\ln =0\]

Answer 1

Hint: As we know that the sum of roots and product of roots of a quadratic equation \[a{{x}^{2}}+bx+c=0\] is given as follows:
Sum of roots \[=\dfrac{-Coefficient\text{ }of\text{ }x}{Coefficient\text{ }of\text{ }{{x}^{2}}}=\dfrac{-b}{a}\]
Product of roots \[=\dfrac{Constant\text{ }term}{Coefficient\text{ }of\text{ }{{x}^{2}}}=\dfrac{c}{a}\]

Complete step-by-step answer:
So we will use the formula and by eliminating \[\sin \alpha \] and \[\cos \alpha \] using trigonometric identity we will get the required result.
We have been given that \[\sin \alpha \] and \[\cos \alpha \] are the roots of the equation \[l{{x}^{2}}+mx+n=0\].
As we know that the sum of roots and product of roots of a quadratic equation is as follows:
Sum of roots \[=\dfrac{-Coefficient\text{ }of\text{ }x}{Coefficient\text{ }of\text{ }{{x}^{2}}}\]
Product of roots \[=\dfrac{Constant\text{ }term}{Coefficient\text{ }of\text{ }{{x}^{2}}}\]
Now we have a quadratic equation \[l{{x}^{2}}+mx+n=0\] whose roots are \[\sin \alpha \] and \[\cos \alpha \] .
Sum of roots \[=\sin \alpha +\cos \alpha =\dfrac{-m}{l}\]
\[sin\alpha +\cos \alpha =\dfrac{-m}{l}.....(1)\]
Product of roots \[=\sin \alpha \cos \alpha =\dfrac{n}{l}....(2)\]
Now on squaring equation (1) we get as follows:
\[\begin{align}
  & {{\left( \sin \alpha +\cos \alpha \right)}^{2}}={{\left( \dfrac{-m}{l} \right)}^{2}} \\
 & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +2\sin \alpha \cos \alpha =\dfrac{{{m}^{2}}}{{{l}^{2}}} \\
\end{align}\]
We know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. So by using this identity in the above equation, we get as follows:
\[\begin{align}
  & \left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \right)+2\sin \alpha \cos \alpha =\dfrac{{{m}^{2}}}{{{l}^{2}}} \\
 & 1+2\sin \alpha \cos \alpha =\dfrac{{{m}^{2}}}{{{l}^{2}}} \\
\end{align}\]
Also, we have \[\sin \alpha \cos \alpha =\dfrac{n}{l}\] from equation (2). So by substituting his value in the above equation we get as follows:
\[\begin{align}
  & 1+2\left( \sin \alpha \cos \alpha \right)=\dfrac{{{m}^{2}}}{{{l}^{2}}} \\
 & 1+2\left( \dfrac{n}{l} \right)=\dfrac{{{m}^{2}}}{{{l}^{2}}} \\
\end{align}\]
On multiplying the above equation by ‘\[{{l}^{2}}\]’ we get as follows:
\[\begin{align}
  & {{l}^{2}}\left( 1+\dfrac{2n}{l} \right)=\dfrac{{{m}^{2}}}{{{l}^{2}}}\times {{l}^{2}} \\
 & {{l}^{2}}+{{l}^{2}}\times \dfrac{2n}{l}={{m}^{2}} \\
 & {{l}^{2}}+2nl={{m}^{2}} \\
\end{align}\]
On taking the term ‘\[{{m}^{2}}\]’ to left hand side of the equation, we get as follows:
\[{{l}^{2}}-{{m}^{2}}+2nl=0\]
Therefore, the correct answer of the above question is option C.

Note: We can also get the solution of the above question by substituting the roots in the equation and thus eliminating it by using trigonometric identity if we don’t know the formula of sum of roots and product of roots. Be careful of the sign while solving the equation.