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Mathematics Question on Quadratic Equations

Given tan Atan\text{ }A and tan Btan\text{ B} are the roots of x2−ax+b=0{{x}^{2}}-ax+b=0 . The value of sin⁡2(A+B){{\sin }^{2}}(A+B) is

A

a2a2+(1−b)2\frac{{{a}^{2}}}{{{a}^{2}}+{{(1-b)}^{2}}}

B

a2a2+b2\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}

C

a2(a+b)2\frac{{{a}^{2}}}{{{(a+b)}^{2}}}

D

b2a2+(1−b)2\frac{{{b}^{2}}}{{{a}^{2}}+{{(1-b)}^{2}}}

Answer

a2a2+(1−b)2\frac{{{a}^{2}}}{{{a}^{2}}+{{(1-b)}^{2}}}

Explanation

Solution

Given that tan A and tan B are the roots of x2−ax+b=0.{{x}^{2}}-ax+b=0. ∴\therefore tanA+tanB=a and tanA tanB=btanA+tanB=a\text{ }and\text{ }tanA\text{ }tanB=b Now, tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B=a1−b\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{a}{1-b} Now, sin⁡2(A+B)=12[1−cos⁡2(A+B)]{{\sin }^{2}}(A+B)=\frac{1}{2}[1-\cos 2(A+B)] =12[1−1−tan⁡2(A+B)1+tan⁡2(A+B)]=\frac{1}{2}\left[ 1-\frac{1-{{\tan }^{2}}(A+B)}{1+{{\tan }^{2}}(A+B)} \right] =12[1+tan⁡2(A+B)−tan⁡2(A+B)1+tan⁡2(A+B)]=\frac{1}{2}\left[ \frac{1+{{\tan }^{2}}(A+B)-{{\tan }^{2}}(A+B)}{1+{{\tan }^{2}}(A+B)} \right] =[tan⁡2(A+B)1+tan⁡2(A+B)]=a2/(1−b)2[a2+(1−b)2](1−b)2=\left[ \frac{{{\tan }^{2}}(A+B)}{1+{{\tan }^{2}}(A+B)} \right]=\frac{{{a}^{2}}/{{(1-b)}^{2}}}{\frac{[{{a}^{2}}+{{(1-b)}^{2}}]}{{{(1-b)}^{2}}}} =a2a2+(1−b)2=\frac{{{a}^{2}}}{{{a}^{2}}+{{(1-b)}^{2}}}