Question

Choose the correct option according to the information. If \[\sec \alpha \] and \[\csc \alpha \] are the roots of the equation \[{x^2} - px + q = 0\].
A) \[{p^2} = q\left( {q - 2} \right)\]
B) \[{p^2} = q\left( {q + 2} \right)\]
C) \[{p^2} + {q^2} = 2q\]
D) None of these

Answer 1

Hint: Here we are given two roots of the equation so we will use the relation between the roots of the quadratic equation to get the desired answer.
The relation is given by :
\[
  \alpha + \beta = \dfrac{{ - b}}{a} \\
  \alpha \beta = \dfrac{c}{a} \\
 \]
Where and are the roots of the quadratic equation

Complete step by step solution:
The given quadratic equation is:-
\[{x^2} - px + q = 0\]
Also the roots of the equation are:-
\[\sec \alpha \] and \[\csc \alpha \]
Now as we know that for standard quadratic equation \[a{x^2} + bx + c = 0\] if \[\alpha \] and \[\beta \] are the roots then,
\[
  \alpha + \beta = \dfrac{{ - b}}{a} \\
  \alpha \beta = \dfrac{c}{a} \\
 \]
Applying this concept for the given equation we get:-
\[
  \sec \alpha + \csc \alpha = \dfrac{{ - \left( { - p} \right)}}{1} \\
  \sec \alpha + \csc \alpha = p................\left( 1 \right) \\
  \sec \alpha \csc \alpha = \dfrac{q}{1} \\
  \sec \alpha \csc \alpha = q...................\left( 2 \right) \\
 \]
Where \[a = 1,b = - p\] and \[c = q\]
Now since we know that:
\[
  \sec \theta = \dfrac{1}{{\cos \theta }} \\
  \csc \theta = \dfrac{1}{{\sin \theta }} \\
 \]
Therefore replacing these values in equation 1 we get:
\[
  \dfrac{1}{{\cos \alpha }} + \dfrac{1}{{\sin \alpha }} = p \\
  \dfrac{{\sin \alpha + \cos \alpha }}{{\sin \alpha \cos \alpha }} = p \\
 \]
Squaring both the sides of the equation:
\[{\left( {\dfrac{{\sin \alpha + \cos \alpha }}{{\sin \alpha \cos \alpha }}} \right)^2} = {p^2}\]
Now, applying the following formula:
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
we get:
\[\dfrac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha + 2\sin \alpha \cos \alpha }}{{{{\sin }^2}\alpha {{\cos }^2}\alpha }} = {p^2}\]
Also we know that:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Therefore,
\[\dfrac{{1 + 2\sin \alpha \cos \alpha }}{{{{\sin }^2}\alpha {{\cos }^2}\alpha }} = {p^2}..................\left( 3 \right)\]
Now, replacing the values of \[\sec \alpha \] and \[\csc \alpha \] in equation 2 we get:
\[
  \dfrac{1}{{\cos \alpha \sin \alpha }} = q \\
  \cos \alpha \sin \alpha = \dfrac{1}{q}.........................\left( 4 \right) \\
 \]
Putting the value of equation 4 in equation 3 we get:
\[
  \dfrac{{1 + 2\left( {\dfrac{1}{q}} \right)}}{{{{\left( {\dfrac{1}{q}} \right)}^2}}} = {p^2} \\
  1 + 2\left( {\dfrac{1}{q}} \right) = {p^2}{\left( {\dfrac{1}{q}} \right)^2} \\
  \dfrac{{q + 2}}{q} = \dfrac{{{p^2}}}{{{q^2}}} \\
  q + 2 = \dfrac{{{p^2}}}{q} \\
  {p^2} = q\left( {q + 2} \right) \\
 \]

Therefore, option B is the correct option.

Note:
We have to use the relation between the roots of the quadratic equation to solve this question.
The trigonometric identities used are:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Also, we have used an algebraic identity:
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]