Question

Given that Tan A, Tan B are the roots of the equation $x^2-bx-c=0$ then
the value of ${\sin }^2\left(A+B\right)$ is

• A. $\frac{b}{{\left(b+c\right)}^2}$
• B. $\frac{b^2}{b^2+c^2}$
• C. $\frac{b^2}{c^2+{\left(1-b\right)}^2}$
• D. $\frac{b^2}{b^2+{\left(1-c\right)}^2}$

Answer 1

The correct option is D $\frac{b^2}{b^2+{\left(1-c\right)}^2}$

$\Rightarrow x^2-bx-c=0$

roots of this equation will be given as

$x=\frac{-b\pm \sqrt{b^2+4c}}{2}$

say $\tan A=\frac{-b+\sqrt{b^2+4c}}{2}\tan B=\frac{-b-\sqrt{b^2+4c}}{2}$

$\therefore \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A+\tan B}$

$=\frac{\frac{-b\sqrt{b^2+4c}}{2}+\frac{-b-\sqrt{b^2+4c}}{2}}{1-\frac{-b+\sqrt{b^2+4c}}{2},\frac{-b-\sqrt{b^2+4c}}{2}}$

$=\frac{\frac{-b+\sqrt{b^2+4c}-b-\sqrt{b^2+4c}}{2}}{\frac{4-\left[\right(-b{)}^2]-(\sqrt{b^2+4c}{)}^2}{2\times 2}}$

$=\frac{-2b\times 2}{4-b^2+b^2-4c}$

$\tan (A+B)=\frac{-4b}{-4c+4}=\frac{-b}{1-c}$

Now, $\Rightarrow \sin (A+B)$

$\Rightarrow \frac{1}{2}{\sin }^2(A+B)=\frac{1}{2}[1-\cos 2(A+B\left)\right]$

$\Rightarrow \frac{1}{2}[1-\frac{1-{\tan }^2(A+B)}{1+{\tan }^2(A+B)}]$

$\Rightarrow \frac{1}{2}[1-\frac{1-\frac{b^2}{(1-c{)}^2}}{1+\frac{b^2}{(1-c{)}^2}}]$

$\Rightarrow \frac{1}{2}[1-\frac{(1-c{)}^2-b^2}{(1-c{)}^2+b^2}]$

$\Rightarrow \frac{1}{2}\left[\frac{(1-c{)}^2+b^2-(1-c{)}^2+b^2}{(1-c{)}^2+b^2}\right]$

$\Rightarrow \frac{1}{2}\times \frac{2b^2}{(1-c{)}^2+b^2}=\frac{b^2}{(1-c{)}^2+b^2}$

$\Rightarrow \frac{b^2}{b^2+(1-c)}$