Question

Given that \[\tan \alpha \]and \[\tan \beta \]are the roots of \[\begin{array}{*{20}{c}}
  {{x^2} - px + q}& = &0
\end{array}\], then the value of \[{\sin ^2}\left( {\alpha + \beta } \right)\]
A) \[\dfrac{{{p^2}}}{{{p^2} + {{\left( {1 - q} \right)}^2}}}\]
B) \[\dfrac{{{p^2}}}{{{p^2} + {q^2}}}\]
C) \[\dfrac{{{q^2}}}{{{p^2} + {{\left( {1 - q} \right)}^2}}}\]
D) \[\dfrac{{{p^2}}}{{{{\left( {p + q} \right)}^2}}}\]

Answer 1

Hint: In this question, we will have to determine the value of \[{\sin ^2}\left( {\alpha + \beta } \right)\]. first of all, we will determine the sum and the product of the roots of the equation. And then we will use some trigonometric formulas to achieve the answer. Hence, we will get a suitable answer.



Formula Used:1) \[\begin{array}{*{20}{c}}
  {\tan \left( {\alpha + \beta } \right)}& = &{\dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}}
\end{array}\]
2) \[2\begin{array}{*{20}{c}}
  {{{\sin }^2}\theta }& = &{1 - \cos 2\theta }
\end{array}\]
3) \[\begin{array}{*{20}{c}}
  {\cos 2\theta }& = &{\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}}
\end{array}\]



Complete step by step solution:According to the question, we have the equation whose roots are given as \[\tan \alpha \]and \[\tan \beta \]. Therefore, we can write it as,
\[\begin{array}{*{20}{c}}
  { \Rightarrow {x^2} - px + q}& = &0
\end{array}\]
Now we will determine the sum and the product of the roots of the given equation. Therefore,
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\tan \alpha + \tan \beta }& = &p
\end{array}\] ------ (1)
And
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\tan \alpha \tan \beta }& = &q
\end{array}\] -------- (2)
Now we will have to determine the value of the \[{\sin ^2}\left( {\alpha + \beta } \right)\]. Therefore, for that purpose
\[\begin{array}{*{20}{c}}
  { \Rightarrow {{\sin }^2}\left( {\alpha + \beta } \right)}& = &{\dfrac{{1 - \cos 2\left( {\alpha + \beta } \right)}}{2}}
\end{array}\]
Now we know the formula of the \[\cos 2\left( {\alpha + \beta } \right)\]. Therefore,
\[\begin{array}{*{20}{c}}
  { \Rightarrow {{\sin }^2}\left( {\alpha + \beta } \right)}& = &{\dfrac{1}{2}\left\{ {1 - \dfrac{{1 - {{\tan }^2}\left( {\alpha + \beta } \right)}}{{1 + {{\tan }^2}\left( {\alpha + \beta } \right)}}} \right\}}
\end{array}\] ……………. (a)
Now to determine the value of the \[\tan \left( {\alpha + \beta } \right)\], we will do
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\tan \left( {\alpha + \beta } \right)}& = &{\dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}}
\end{array}\]
Now put the value of the equation (1) and (2) in the above formula
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\tan \left( {\alpha + \beta } \right)}& = &{\dfrac{p}{{1 - q}}}
\end{array}\] …………… (3)
Now we will put the value of equation (3) in equation (a). Therefore, we will get
\[\begin{array}{*{20}{c}}
  { \Rightarrow {{\sin }^2}\left( {\alpha + \beta } \right)}& = &{\dfrac{1}{2}\left\{ {1 - \dfrac{{1 - {{\left( {\dfrac{p}{{1 - q}}} \right)}^2}}}{{1 + {{\left( {\dfrac{p}{{1 - q}}} \right)}^2}}}} \right\}}
\end{array}\]
Now we will simplify the above expression. Therefore,
\[\begin{array}{*{20}{c}}
  { \Rightarrow {{\sin }^2}\left( {\alpha + \beta } \right)}& = &{\dfrac{1}{2}\left\{ {\dfrac{{{{\left( {1 - q} \right)}^2} + {p^2} - {{\left( {1 - q} \right)}^2} + {p^2}}}{{{{\left( {1 - q} \right)}^2} + {p^2}}}} \right\}}
\end{array}\]
Finally, we will get,
\[\begin{array}{*{20}{c}}
  { \Rightarrow {{\sin }^2}\left( {\alpha + \beta } \right)}& = &{\dfrac{{{p^2}}}{{{{\left( {1 - q} \right)}^2} + {p^2}}}}
\end{array}\]
Hence, we can choose the correct answer from the given option.



Option ‘A’ is correct

Note: In this question, the first point is to keep in mind that change the \[{\sin ^2}\left( {\alpha + \beta } \right)\]in the form of \[\tan \left( {\alpha + \beta } \right)\]. Hence, we will be able to find the desired answer.