Question

If $\tan A$ and $\tan B$ are the roots of the equation $x^2-px+q=0$, then the value of ${\sin }^2(A+B)$ is

• A. $\frac{p^2}{p^2+{(1-q)}^2}$
• B. $\frac{p^2}{p^2-{(1-q)}^2}$
• C. $\frac{q^2}{p^2+{(1-q)}^2}$
• D. None of these

Answer 1

Answer: (A)

Given:

$\tan A,\tan B$ are the roots of $x^2-px+q=0$.

$\Rightarrow \tan A+\tan B=p,\tan A.\tan B=q$

Now, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{p}{1-q}$ ---(1)

We know that, ${\sin }^2\left(\theta \right)=\frac{1-\cos 2\left(\theta \right)}{2}$

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

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

$=\frac{{\tan }^2(A+B)}{1+{\tan }^2(A+B)}$

$=\frac{{\left[\frac{p}{(1-q)}\right]}^2}{1+{\left[\frac{p}{(1-q)}\right]}^2}$

$=\frac{p^2}{{(1-q)}^2+p^2}$

$\therefore {\sin }^2(A+B)=\frac{p^2}{{(1-q)}^2+p^2}$

Hence, Option A is correct.