Question

If $sec\left(\alpha \right)$ and $\cos ec\left(\alpha \right)$ are the roots of the equation $x^2-px+q=0$, then prove that $p^2=q(q+2)$.

Answer 1

Prove the given statement.

If a quadratic equation is in the form of $a{x}^2+bx+c=0$.

Then, The sum and product of roots of a quadratic equation can be given by $-\frac{b}{a}$ and $\frac{c}{a}$ respectively. Where $a,b$ are the coefficients of $x^2,x$ respectively.

Since, $sec\left(\alpha \right)$ and $\cos ec\left(\alpha \right)$ are the roots of the equation $x^2-px+q=0$.

Therefore, $sec\left(\alpha \right)+\cos ec\left(\alpha \right)=p$

$$\begin{gathered} \Rightarrow \frac{1}{\cos \left(\alpha \right)}+\frac{1}{\sin \left(\alpha \right)}=p \\ \Rightarrow \frac{\sin \left(\alpha \right)+\cos \left(\alpha \right)}{\sin \left(\alpha \right)\cos \left(\alpha \right)}=p \end{gathered}$$

Take square on both sides.

$$\begin{gathered} \Rightarrow p^2=\frac{{\cos }^2\left(\alpha \right)+{\sin }^2\left(\alpha \right)+2\cos \left(\alpha \right)\sin \left(\alpha \right)}{{\cos }^2\left(\alpha \right){\sin }^2\left(\alpha \right)} \\ \Rightarrow p^2=\frac{1+2\cos \left(\alpha \right)\sin \left(\alpha \right)}{{\cos }^2\left(\alpha \right){\sin }^2\left(\alpha \right)} \\ \Rightarrow p^2=\frac{1}{{\cos }^2\left(\alpha \right){\sin }^2\left(\alpha \right)}+\frac{2}{\cos \left(\alpha \right)\sin \left(\alpha \right)}...1 \end{gathered}$$

Also, $sec\left(\alpha \right)\cos ec\left(\alpha \right)=q$

$$\begin{gathered} \Rightarrow sec\left(\alpha \right)\cos ec\left(\alpha \right)=q \\ \Rightarrow \frac{1}{\cos \left(\alpha \right)\sin \left(\alpha \right)}=q...2 \end{gathered}$$

From equation $1$ and equation $2$ we get.

$$\begin{gathered} p^2=q^2+2q \\ \Rightarrow p^2=q\left(q+2\right) \end{gathered}$$

Hence it is proved, $p^2=q(q+2)$.