Question

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

Answer 1

Hint: In this problem it is given that \[\angle R=\dfrac{\pi }{2}\] that means \[\angle P+\angle Q=\dfrac{\pi }{2}\] by using the property of triangle ten we have to use the formula for trigonometry and also we know that sum of roots of a quadratic equation is \[\dfrac{-b}{a}\] then after simplifying and substituting in the formula and get the answer.

Complete step by step solution:
In this type of problem, \[\Delta PQR,\] given that \[\angle R=\dfrac{\pi }{2}\] that means by using the angle sum property that is
\[\angle P+\angle Q+\angle R=\pi \]
By substituting the values we get:
\[\angle P+\angle Q+\dfrac{\pi }{2}=\pi \]
Further solving this we get:
\[\angle P+\angle Q=\pi -\dfrac{\pi }{2}\]
Therefore, we get: \[\angle P+\angle Q=\dfrac{\pi }{2}\]
According to the given question there are two roots of quadratic equation that is \[\tan \left( \dfrac{p}{2} \right)\] and \[\tan \left( \dfrac{Q}{2} \right)\]of\[a{{x}^{2}}+bx+c=0\], \[a\ne 0\]
Sum of the roots of the quadratic equation is given by \[\dfrac{-b}{a}\]
That means \[\tan \left( \dfrac{p}{2} \right)+\tan \left( \dfrac{Q}{2} \right)=\dfrac{-b}{a}\]
By using the property of trigonometry formula we get:
\[\tan \left( A+B \right)=\dfrac{\tan \left( A \right)+\tan \left( B \right)}{1-\tan \left( A \right)\tan \left( B \right)}\]
By applying the formula we get:
\[\dfrac{\tan \left( \dfrac{P}{2} \right)+\tan \left( \dfrac{Q}{2} \right)}{1-\tan \left( \dfrac{P}{2} \right)\tan \left( \dfrac{Q}{2} \right)}=\tan \left( \dfrac{P}{2}+\dfrac{Q}{2} \right)\]
By simplification we get:
\[\dfrac{\tan \left( \dfrac{P}{2} \right)+\tan \left( \dfrac{Q}{2} \right)}{1-\tan \left( \dfrac{P}{2} \right)\tan \left( \dfrac{Q}{2} \right)}=\tan \left( \dfrac{P+Q}{2} \right)\]
As we know that\[\angle P+\angle Q=\dfrac{\pi }{2}\] substituting in the above equation we get:
\[\dfrac{\tan \left( \dfrac{P}{2} \right)+\tan \left( \dfrac{Q}{2} \right)}{1-\tan \left( \dfrac{P}{2} \right)\tan \left( \dfrac{Q}{2} \right)}=\tan \left( \dfrac{\dfrac{\pi }{2}}{2} \right)\]
By simplification we get:
\[\dfrac{\tan \left( \dfrac{P}{2} \right)+\tan \left( \dfrac{Q}{2} \right)}{1-\tan \left( \dfrac{P}{2} \right)\tan \left( \dfrac{Q}{2} \right)}=\tan \left( \dfrac{\pi }{4} \right)\]
As we know that \[\tan \left( \dfrac{\pi }{4} \right)=1\] substitute in the above equation we get:
\[\dfrac{\tan \left( \dfrac{P}{2} \right)+\tan \left( \dfrac{Q}{2} \right)}{1-\tan \left( \dfrac{P}{2} \right)\tan \left( \dfrac{Q}{2} \right)}=1\]
As we know, the Sum of roots of quadratic equation is given by \[\dfrac{-b}{a}\] and product of roots of quadratic equation is given by \[\dfrac{c}{a}\]
That is\[\tan \left( \dfrac{p}{2} \right)+\tan \left( \dfrac{Q}{2} \right)=\dfrac{-b}{a}\] and \[\tan \left( \dfrac{P}{2} \right)\tan \left( \dfrac{Q}{2} \right)=\dfrac{c}{a}\] substitute this value on above equation we get:
\[\dfrac{\dfrac{-b}{a}}{1-\dfrac{c}{a}}=1\]
By simplifying further we get:
\[\dfrac{-b}{a}=1-\dfrac{c}{a}\]
By cross multiplying and further simplifying this we get:
\[-b=a-c\]
By rearranging the term we get:
\[c=a+b\]
Therefore, the correct option is option (C).

Note:
In this type of problems, always remember the formula which we used and the trigonometric formula is very important here and also remember the formula for sum and product of roots of a given quadratic equation. If we know one angle then we find the sum of two angles by using the angle sum property. In a Euclidean space, the sum of angles of a triangle equals $180^\circ$. Whether a triangle is an acute, obtuse, or a right triangle, the sum of the angles will always be $180^\circ$. Thus, the angle sum property states that the sum of the angles of a triangle is equal to $180^\circ$.