Question

If an angle \[A\] of a \[\triangle ABC\] satisfies \[5cosA + 3 = 0\], then the roots of the quadratic equation, \[9x^{2} + 27x + 20 = 0\] are
A \[sin\ A,\ sec\ A\]
B \[sec\ A,\ tan\ A\]
C \[tan\ A,\ cos\ A\]
D \[sec\ A,\ cot\ A\]

Answer 1

Hint: In this question, we need to find the roots of the quadratic equation, \[9x^{2} + 27x + 20 = 0\] in terms of trigonometric terms. First we need to use the quadratic formula to find the roots of the quadratic equation, \[9x^{2} + 27x + 20 = 0\]. Then we need to find the value of cos A by using the given equation \[5cosA+3=0\] . Then by using trigonometric identities, we can find the roots of the equation.

Complete step by step solution: 
Given quadratic equation, \[9x^{2} + 27x + 20 = 0\]
On comparing the given quadratic equation with \[ax^{2} + bx + c = 0\] , we get, \[a = 9\] , \[b = 27\] and \[c = 20\]
We can use the quadratic formula to find the roots of the given quadratic equation.
Quadratic formula :
\[x = \dfrac{- b \pm \sqrt{b^{2} – 4ac}}{2a}\]
On substituting the value of \[a\], \[b\] and \[c\] in the formula,
We get,
\[x = \dfrac{- 27 \pm \sqrt{27^{2} – 4 \times 9 \times 20}}{2 \times 9}\]
On solving,
We get,
\[x = \dfrac{- 27 \pm \sqrt{729 – 720}}{18}\]
On further simplifying,
We get,
\[x = \dfrac{- 27 \pm \sqrt{9}}{18}\]
On expanding,
We get,
\[x = \dfrac{- 27 + 3}{18}\ \] and \[\ x = \dfrac{- 27 – 3}{18}\]
On simplification,
We get,
\[x = - \dfrac{24}{18}\] and \[x = - \dfrac{30}{18}\]
On further simplification,
We get,
\[x = - \dfrac{4}{3}\] and \[x = - \dfrac{5}{3}\]
Also given \[5cosA + 3 = 0\]
On subtracting \[3\] on both sides,
We get
\[5cos\ A = - 3\]
Now on dividing \[5\] on both sides,
We get,
\[cos\ A = - \dfrac{3}{5}\]
We know that the reciprocal of cosine function is secant function.
That is \[sec\ A\ = \dfrac{1}{cos\ A}\]
On substituting the value of \[cos\ A\] ,
We get,
\[sec\ A\ = \dfrac{1}{- \dfrac{3}{5}}\]
On solving,
We get,
\[sec\ A\ = - \dfrac{5}{3}\]
Then \[tan\ A = - \sqrt{sec^{2}A – 1}\]
On substituting the value of \[sec\ A\],
We get,
\[tan\ A = - \sqrt{\left( - \dfrac{5}{3} \right)^{2} – 1}\]
On solving,
We get,
\[tan\ A = - \sqrt{\dfrac{25}{9} – 1}\]
On taking LCM,
We get,
\[tan\ A = - \sqrt{\dfrac{25 – 9}{9}}\]
On simplifying ,
We get,
\[tan\ A = - \sqrt{\dfrac{16}{9}}\]
On taking the numbers out of the square root,
We get ,
\[tan\ A = - \dfrac{4}{3}\]
The negative sign indicates that the angle \[A\] is an obtuse angle.
Since the roots of the equation \[9x^{2} + 27x + 20 = 0\ \] equals the value of \[sec\ A\] and \[tan\ A\] , the roots of the equation are \[sec\ A\] and \[tan\ A\] .
The roots of the equation are \[sec\ A\] and \[tan\ A\] .
Hence the correct option is B.

Note: In order to solve these questions , we should have a strong grip over quadratic equations and trigonometric identities . We should keep in mind that we need to solve the given quadratic equation in order to find the roots of the equation . We can also directly find the roots of the quadratic equation by using the formula \[x = \dfrac{- b \pm \sqrt{b^{2} – 4ac}}{2a}\] .