Question

If $3{x^2} - 10x - 25 = 0$ is a quadratic equation such that it has two roots $\tan A$and $\tan B$ then find out the value of $3{\sin ^2}\left( {A + B} \right) - 10\sin \left( {A + B} \right).\cos \left( {A + B} \right) - 25{\cos ^2}\left( {A + B} \right)$.
A. $ - 10$
B. $10$
C. $ - 25$
D. $25$

Answer 1

Hint: We know that if $\alpha $and $\beta $ are the roots of quadratic equation of $P{x^2} + Qx + R = 0$ then the roots are written as $\alpha = \dfrac{{ - Q + \sqrt {{Q^2} - 4PR} }}{{2P}}$ and $\beta = \dfrac{{ - Q - \sqrt {{Q^2} - 4PR} }}{{2P}}$
So, the sum of the two roots are $\alpha + \beta = \dfrac{{ - Q}}{P}$ and the product of the two roots are $\alpha .\beta = \dfrac{R}{P}$

Complete step-by-step solution:
Here the quadratic equation is given $3{x^2} - 10x - 25 = 0$ and the roots are $\tan A$ and $\tan B$ .
So, the sum of the two roots $\tan A + \tan B = \dfrac{{10}}{3}$
 Product of the two roots $\tan A.\tan B = \dfrac{{ - 25}}{3}$
Now we know, $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}$
Putting the above values we get $ \Rightarrow \dfrac{{\dfrac{{10}}{3}}}{{1 - \left( {\dfrac{{ - 25}}{3}} \right)}} = \dfrac{{\dfrac{{10}}{3}}}{{\dfrac{{28}}{3}}} = \dfrac{5}{{14}}$
Now calculate $
  \tan \left( {A + B} \right) = \dfrac{{\sin \left( {A + B} \right)}}{{\cos \left( {A + B} \right)}} = \dfrac{5}{{14}} \\
   \Rightarrow \sin \left( {A + B} \right) = \cos \left( {A + B} \right) \times \dfrac{5}{{14}} \\
   \Rightarrow {\sin ^2}\left( {A + B} \right) = {\cos ^2}\left( {A + B} \right) \times \dfrac{{25}}{{196}}
$
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Put the value of ${\sin ^2}\left( {A + B} \right)$ in this equation,hence we get
$\therefore $ $
  {\cos ^2}\left( {A + B} \right) \times \dfrac{{25}}{{196}} + {\cos ^2}\left( {A + B} \right) = 1 \\
   \Rightarrow \cos \left( {A + B} \right) = \dfrac{{14}}{{\sqrt {221} }} \\
   \Rightarrow \sin \left( {A + B} \right) = \dfrac{5}{{14}} \times \dfrac{{14}}{{\sqrt {221} }} = \dfrac{5}{{\sqrt {221} }}
$
Now put the value of $\sin \left( {A + B} \right)$ and $\cos \left( {A + B} \right)$ in the following equation,
$3{\sin ^2}\left( {A + B} \right) - 10\sin \left( {A + B} \right).\cos \left( {A + B} \right) - 25{\cos ^2}\left( {A + B} \right)$
$
 = 3 \times \dfrac{{25}}{{221}} - 10 \times \dfrac{5}{{\sqrt {221} }} \times \dfrac{{14}}{{\sqrt {221} }} - 25 \times \dfrac{{196}}{{221}} \\
 = \dfrac{{75 - 700 - 4900}}{{221}} \\
 = - 25
$
The value of the required expression is $ - 25$

Hence, option (C) will be the right answer.

Note: Here we used $P{x^2} + Qx + R = 0$ as a quadratic equation because A and B are used in the question.
Not all equations can be easily factored. Thus we need a formula to solve the equation so we use a quadratic formula. There are three conditions –
1. If ${Q^2} - 4PR\rangle 0$ then we will get two real solutions.
2. If ${Q^2} - 4PR = 0$ then we will get a double root.
3. If ${Q^2} - 4PR\langle 0$ then we will get two complex solutions.