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Question: If \(3{x^2} - 10x - 25 = 0\) is a quadratic equation such that it has two roots \(\tan A\)and \(\tan...

If 3x2−10x−25=03{x^2} - 10x - 25 = 0 is a quadratic equation such that it has two roots tan⁡A\tan Aand tan⁡B\tan B then find out the value of 3sin⁡2(A+B)−10sin⁡(A+B).cos⁡(A+B)−25cos⁡2(A+B)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 - 10
B. 1010
C. −25 - 25
D. 2525

Explanation

Solution

We know that if α\alpha and β\beta are the roots of quadratic equation of Px2+Qx+R=0P{x^2} + Qx + R = 0 then the roots are written as α=−Q+Q2−4PR2P\alpha = \dfrac{{ - Q + \sqrt {{Q^2} - 4PR} }}{{2P}} and β=−Q−Q2−4PR2P\beta = \dfrac{{ - Q - \sqrt {{Q^2} - 4PR} }}{{2P}}
So, the sum of the two roots are α+β=−QP\alpha + \beta = \dfrac{{ - Q}}{P} and the product of the two roots are α.β=RP\alpha .\beta = \dfrac{R}{P}

Complete step-by-step solution:
Here the quadratic equation is given 3x2−10x−25=03{x^2} - 10x - 25 = 0 and the roots are tan⁡A\tan A and tan⁡B\tan B .
So, the sum of the two roots tan⁡A+tan⁡B=103\tan A + \tan B = \dfrac{{10}}{3}
Product of the two roots tan⁡A.tan⁡B=−253\tan A.\tan B = \dfrac{{ - 25}}{3}
Now we know, tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡A.tan⁡B\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}
Putting the above values we get ⇒1031−(−253)=103283=514 \Rightarrow \dfrac{{\dfrac{{10}}{3}}}{{1 - \left( {\dfrac{{ - 25}}{3}} \right)}} = \dfrac{{\dfrac{{10}}{3}}}{{\dfrac{{28}}{3}}} = \dfrac{5}{{14}}
Now calculate tan⁡(A+B)=sin⁡(A+B)cos⁡(A+B)=514 ⇒sin⁡(A+B)=cos⁡(A+B)×514 ⇒sin⁡2(A+B)=cos⁡2(A+B)×25196 \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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
Put the value of sin⁡2(A+B){\sin ^2}\left( {A + B} \right) in this equation,hence we get
∴\therefore cos⁡2(A+B)×25196+cos⁡2(A+B)=1 ⇒cos⁡(A+B)=14221 ⇒sin⁡(A+B)=514×14221=5221 {\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⁡(A+B)\sin \left( {A + B} \right) and cos⁡(A+B)\cos \left( {A + B} \right) in the following equation,
3sin⁡2(A+B)−10sin⁡(A+B).cos⁡(A+B)−25cos⁡2(A+B)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)
=3×25221−10×5221×14221−25×196221 =75−700−4900221 =−25 = 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 - 25

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

Note: Here we used Px2+Qx+R=0P{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 Q2−4PR⟩0{Q^2} - 4PR\rangle 0 then we will get two real solutions.
2. If Q2−4PR=0{Q^2} - 4PR = 0 then we will get a double root.
3. If Q2−4PR⟨0{Q^2} - 4PR\langle 0 then we will get two complex solutions.