Question

If $\sin \left(\theta \right)+\cos \left(\theta \right)=h$, then the quadratic equation having $\sin \left(\theta \right)$ and $\cos \left(\theta \right)$ as its roots, is

• A. $x^2–hx+(h^2–1)=0$
• B. $2x^2–2hx+(h^2–1)=0$
• C. $x^2–hx+2(h^2–1)=0$
• D. $x^2–2hx+(h^2–1)=0$

Answer 1

The correct option is B

$2x^2–2hx+(h^2–1)=0$

Explanation for the correct option:

Step 1: find the relation between roots

The quadratic equation is an equation of the form $\mathit{a}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{x}\mathbf{+}\mathit{c}\mathbf{=}\mathbf{0}$,

The sum of the roots of the above equation is calculated by the formula

Sum of the roots $\mathbf{=}\mathbf{-}\frac{\mathit{b}}{\mathit{a}}$

The product of the roots is calculated by the formula,

Product of the roots $\mathbf{=}\frac{\mathit{c}}{\mathit{a}}$

Step 2: Determination of the quadratic equation

Here, the roots of the equation are $\sin \left(\theta \right)$ and $\cos \left(\theta \right)$. Also, the sum of the roots is given which is

$\sin \left(\theta \right)+\cos \left(\theta \right)=h$

To find the product of the roots square both the sides of the above equation,

$\begin{array}{rcl}\therefore {\left(\sin \left(\theta \right)+\cos \left(\theta \right)\right)}^2& =& h^2\\ & \Rightarrow & {\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)+2\sin \left(\theta \right)\cos \left(\theta \right)=h^2\\ & \Rightarrow & 1+2s\mathrm{in}\left(\theta \right)\cos \left(\theta \right)=h^2\\ & \Rightarrow & \sin \left(\theta \right)\cos \left(\theta \right)=\frac{h^2-1}{2}\end{array}$

Now, the quadratic equation can be written as,

$x^2-hx+\frac{h^2-1}{2}=0$

Multiply both the sides by $2$, we get

$2x^2-2hx+\left(h^2-1\right)=0$

Hence, the correct option is (B).