Question

If \[\tan\alpha \], and \[\tan\beta \] are the roots of the equation \[{x^2} + px + q = 0\] where \[\left( {p \ne 0} \right)\] , then which of the following equation is true?
A. \[\sin^{2}\left( {\alpha + \beta } \right) + p\sin\left( {\alpha + \beta } \right)\cos\left( {\alpha + \beta } \right) + q\cos^{2}\left( {\alpha + \beta } \right) = q\]
B. \[\tan\left( {\alpha + \beta } \right) = \dfrac{p}{{q - 1}}\]
C. \[\cos\left( {\alpha + \beta } \right) = 1 - q\]
D. \[\sin\left( {\alpha + \beta } \right) = - p\]

Answer 1

Hint: First, calculate the sum and product of the roots of a quadratic equation. Then substitute the values of the sum and product of the roots in the formula of \[\tan\left( {\alpha + \beta } \right)\] and solve it. After that, use the trigonometric ratio \[\tan A = \dfrac{{\sin A}}{{\cos A}}\] to calculate the values of \[\sin\left( {\alpha + \beta } \right)\], and \[\cos\left( {\alpha + \beta } \right)\]. In the end, substitute the values of \[\sin\left( {\alpha + \beta } \right)\], and \[\cos\left( {\alpha + \beta } \right)\] in the expression \[\sin^{2}\left( {\alpha + \beta } \right) + p\sin\left( {\alpha + \beta } \right)\cos\left( {\alpha + \beta } \right) + q\cos^{2}\left( {\alpha + \beta } \right)\] and find the value of the expression. Verify the values of each expression with the given options to get the required answer.

Formula used
\[\tan A = \dfrac{{\sin A}}{{\cos A}}\]
\[\tan\left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A \tan B}}\]
\[\sin^{2}A + \cos^{2}A = 1\]

Complete step by step solution
Given:
\[\tan\alpha \], and \[\tan\beta \] are the roots of the equation \[{x^2} + px + q = 0\] where \[\left( {p \ne 0} \right)\]

Let’s calculate the sum and product of the roots of the given quadratic equation.
We get,
Sum of the roots: \[\tan \alpha + \tan \beta = - p\]
Product of the roots: \[\tan \alpha \times \tan \beta = q\]

Now verify each option.
Substitute the values of \[\tan\alpha + \tan\beta = - p\], and \[\tan \alpha \times \tan \beta = q\] in the formula of \[\tan\left( {\alpha + \beta } \right)\].
We get,
\[\tan\left( {\alpha + \beta } \right) = \dfrac{{tan \alpha + \tan \beta }}{{1 - \tan \alpha \times tan \beta }}\]
\[ \Rightarrow \tan\left( {\alpha + \beta } \right) = \dfrac{{ - p}}{{1 - q}}\]
\[ \Rightarrow \tan\left( {\alpha + \beta } \right) = \dfrac{p}{{q - 1}}\]
Hence, option B is correct.

Now apply the trigonometric ratio \[\tan A = \dfrac{{\sin A}}{{\cos A}}\] to find the values of \[\sin\left( {\alpha + \beta } \right)\], and \[\cos\left( {\alpha + \beta } \right)\].
We get,
\[\tan \left( {\alpha + \beta } \right) = \dfrac{{\sin\left( {\alpha + \beta } \right)}}{{\cos \left( {\alpha + \beta } \right)}}\]
\[ \Rightarrow \dfrac{p}{{q - 1}} = \dfrac{{\sin\left( {\alpha + \beta } \right)}}{{\cos \left( {\alpha + \beta } \right)}}\]
After equating both sides. We get
\[\sin\left( {\alpha + \beta } \right) = p\] and \[\cos\left( {\alpha + \beta } \right) = q - 1\]
From the above equations, we observe that the options C and D are incorrect.

Now substitute the values of \[\sin\left( {\alpha + \beta } \right)\], and \[\cos\left( {\alpha + \beta } \right)\] in the expression \[\sin^{2}\left( {\alpha + \beta } \right) + p\sin\left( {\alpha + \beta } \right)\cos\left( {\alpha + \beta } \right) + q\cos^{2}\left( {\alpha + \beta } \right)\].
\[\sin^{2}\left( {\alpha + \beta } \right) + p\sin\left( {\alpha + \beta } \right)\cos\left( {\alpha + \beta } \right) + q\cos^{2}\left( {\alpha + \beta } \right) = {p^2} + p \times p \times \left( {q - 1} \right) + q{\left( {q - 1} \right)^2}\]
\[ \Rightarrow \sin^{2}\left( {\alpha + \beta } \right) + p\sin\left( {\alpha + \beta } \right)\cos\left( {\alpha + \beta } \right) + q\cos^{2}\left( {\alpha + \beta } \right) = {p^2} + {p^2}\left( {q - 1} \right) + q{\left( {q - 1} \right)^2}\]
\[ \Rightarrow \sin^{2}\left( {\alpha + \beta } \right) + p\sin\left( {\alpha + \beta } \right)\cos\left( {\alpha + \beta } \right) + q\cos^{2}\left( {\alpha + \beta } \right) = {p^2} + {p^2}q - {p^2} + q{\left( {q - 1} \right)^2}\]
\[ \Rightarrow \sin^{2}\left( {\alpha + \beta } \right) + p\sin\left( {\alpha + \beta } \right)\cos\left( {\alpha + \beta } \right) + q\cos^{2}\left( {\alpha + \beta } \right) = {p^2}q + q{\left( {q - 1} \right)^2}\]
\[ \Rightarrow \sin^{2}\left( {\alpha + \beta } \right) + p\sin\left( {\alpha + \beta } \right)\cos\left( {\alpha + \beta } \right) + q\cos^{2}\left( {\alpha + \beta } \right) = q\left[ {{p^2} + {{\left( {q - 1} \right)}^2}} \right]\]
\[ \Rightarrow \sin^{2}\left( {\alpha + \beta } \right) + p\sin\left( {\alpha + \beta } \right)\cos\left( {\alpha + \beta } \right) + q\cos^{2}\left( {\alpha + \beta } \right) = q\left[ {\sin^{2}\left( {\alpha + \beta } \right) + \cos^{2}\left( {\alpha + \beta } \right)} \right]\]
\[ \Rightarrow \sin^{2}\left( {\alpha + \beta } \right) + p\sin\left( {\alpha + \beta } \right)\cos\left( {\alpha + \beta } \right) + q\cos^{2}\left( {\alpha + \beta } \right) = q\left[ 1 \right]\]
\[ \Rightarrow \sin^{2}\left( {\alpha + \beta } \right) + p\sin\left( {\alpha + \beta } \right)\cos\left( {\alpha + \beta } \right) + q\cos^{2}\left( {\alpha + \beta } \right) = q\]
Therefore, option A is correct.

Hence the correct options are A and B.



Note: The sum and product of a quadratic equation are calculated by using the coefficients of the variables.
If \[\alpha \], and \[\beta \] are the roots of the general quadratic equation \[a{x^2} + bx + c = 0\] where \[\left( {b \ne 0} \right)\], then
Sum of the roots: \[\alpha + \beta = \dfrac{{ - b}}{a}\]
Product of the roots: \[\alpha \beta = \dfrac{c}{a}\]