Question

In a triangle PQR, \[\angle R = \dfrac{\pi }{2}\]. If \[\tan \dfrac{P}{2}\] and \[\tan \dfrac{Q}{2}\] are the roots of the equation \[a{x^2} + bx + c = 0\left( {a \ne 0} \right)\], then which of the following is true?
A. \[a + b = c\]
B. \[b + c = a\]
C. \[a + c = b\]
D. \[b = c\]

Answer 1

Hint: We will apply the formula sum of roots and product of roots on the given equation. Then simplify these two equations by the property of the angles of the triangle and the tangent of difference between the two angles and solve these two to get the desired result.

Formula used:
The sum of roots of a quadratic equation \[A{x^2} + Bx + C = 0\] is \[ - \dfrac{B}{A}\].
The product of roots of a quadratic equation \[A{x^2} + Bx + C = 0\] is \[\dfrac{C}{A}\].
The formula of the tangent of difference between two angles:
\[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]

Complete step by step solution:
Given equation is \[a{x^2} + bx + c = 0\left( {a \ne 0} \right)\]
The sum of the two roots of the equation is \[ - \dfrac{b}{a}\].
The product of the two roots of the equation is \[\dfrac{c}{a}\].
Given that, \[\tan \dfrac{P}{2}\] and \[\tan \dfrac{Q}{2}\] are the roots of the equation \[a{x^2} + bx + c = 0\left( {a \ne 0} \right)\].
Therefore, \[\tan \dfrac{P}{2} + \tan \dfrac{Q}{2} = - \dfrac{b}{a}\] ….(i)
\[\tan \dfrac{P}{2}\tan \dfrac{Q}{2} = \dfrac{c}{a}\] …..(ii)
We know that the sum of all angles of triangle is \[\pi \].
Thus,\[P + Q + R = \pi \]
Substitute \[\angle R = \dfrac{\pi }{2}\] in the above equation:
\[ \Rightarrow P + Q + \dfrac{\pi }{2} = \pi \]
\[ \Rightarrow Q = \pi - \dfrac{\pi }{2} - P\]
\[ \Rightarrow Q = \dfrac{\pi }{2} - P\]
Substitute \[Q = \dfrac{\pi }{2} - P\]in equation (i)
\[\tan \dfrac{P}{2} + \tan \dfrac{{\dfrac{\pi }{2} - P}}{2} = - \dfrac{b}{a}\]
\[ \Rightarrow \tan \dfrac{P}{2} + \tan \left( {\dfrac{\pi }{4} - \dfrac{P}{2}} \right) = - \dfrac{b}{a}\]
Now applying the formula of the tangent of difference between two angles:
\[ \Rightarrow \tan \dfrac{P}{2} + \dfrac{{\tan \dfrac{\pi }{4} - \tan \dfrac{P}{2}}}{{1 + \tan \dfrac{\pi }{4}\tan \dfrac{P}{2}}} = - \dfrac{b}{a}\]
\[ \Rightarrow \tan \dfrac{P}{2} + \dfrac{{1 - \tan \dfrac{P}{2}}}{{1 + 1 \cdot \tan \dfrac{P}{2}}} = - \dfrac{b}{a}\]
\[ \Rightarrow \dfrac{{\tan \dfrac{P}{2}\left( {1 + \tan \dfrac{P}{2}} \right) + 1 - \tan \dfrac{P}{2}}}{{1 + \tan \dfrac{P}{2}}} = - \dfrac{b}{a}\]
\[ \Rightarrow \dfrac{{\tan \dfrac{P}{2} + {{\tan }^2}\dfrac{P}{2} + 1 - \tan \dfrac{P}{2}}}{{1 + \tan \dfrac{P}{2}}} = - \dfrac{b}{a}\]
\[ \Rightarrow \dfrac{{{{\tan }^2}\dfrac{P}{2} + 1}}{{1 + \tan \dfrac{P}{2}}} = - \dfrac{b}{a}\] …..(iii)
Now we will simplify equation (ii)
\[\tan \dfrac{P}{2}\tan \dfrac{Q}{2} = \dfrac{c}{a}\]
Substitute \[Q = \dfrac{\pi }{2} - P\]in above equation:
\[\tan \dfrac{P}{2}\tan \dfrac{{\dfrac{\pi }{2} - P}}{2} = \dfrac{c}{a}\]
\[ \Rightarrow \tan \dfrac{P}{2}\tan \dfrac{{\dfrac{\pi }{2} - P}}{2} = \dfrac{c}{a}\]
\[ \Rightarrow \tan \dfrac{P}{2}\tan \left( {\dfrac{\pi }{4} - \dfrac{P}{2}} \right) = \dfrac{c}{a}\]
Now applying the formula of the tangent of difference between two angles:
\[ \Rightarrow \tan \dfrac{P}{2}\dfrac{{\tan \dfrac{\pi }{4} - \tan \dfrac{P}{2}}}{{1 + \tan \dfrac{\pi }{4}\tan \dfrac{P}{2}}} = \dfrac{c}{a}\]
\[ \Rightarrow \tan \dfrac{P}{2} \cdot \dfrac{{1 - \tan \dfrac{P}{2}}}{{1 + \tan \dfrac{P}{2}}} = \dfrac{c}{a}\]
\[ \Rightarrow \dfrac{{\tan \dfrac{P}{2} - {{\tan }^2}\dfrac{P}{2}}}{{1 + \tan \dfrac{P}{2}}} = \dfrac{c}{a}\]….(iv)
Add equation (iii) and (iv)
\[\dfrac{{{{\tan }^2}\dfrac{P}{2} + 1}}{{1 + \tan \dfrac{P}{2}}} + \dfrac{{\tan \dfrac{P}{2} - {{\tan }^2}\dfrac{P}{2}}}{{1 + \tan \dfrac{P}{2}}} = - \dfrac{b}{a} + \dfrac{c}{a}\]
\[ \Rightarrow \dfrac{{{{\tan }^2}\dfrac{P}{2} + 1 + \tan \dfrac{P}{2} - {{\tan }^2}\dfrac{P}{2}}}{{1 + \tan \dfrac{P}{2}}} = - \dfrac{b}{a} + \dfrac{c}{a}\]
\[ \Rightarrow \dfrac{{1 + \tan \dfrac{P}{2}}}{{1 + \tan \dfrac{P}{2}}} = - \dfrac{b}{a} + \dfrac{c}{a}\]
\[ \Rightarrow 1 = - \dfrac{b}{a} + \dfrac{c}{a}\]
\[ \Rightarrow c - b = a\]
\[ \Rightarrow a + b = c\]
Hence option A is the correct option.

Note: Students often mistake to apply the sum of roots and product of roots. They take the sum of roots is \[ - \dfrac{b}{c}\] and product of roots is \[\dfrac{a}{c}\] which are incorrect formula. The correct formula are the sum of roots is \[ - \dfrac{b}{a}\] and product of roots is \[\dfrac{c}{a}\].