Question

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

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

Answer 1

The Correct Option is A

Given that tan A and tan B are the roots of $ {{x}^{2}}-ax+b=0. $ $ \therefore $ $ tanA+tanB=a\text{ }and\text{ }tanA\text{ }tanB=b $ Now, $ \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{a}{1-b} $ Now, $ {{\sin }^{2}}(A+B)=\frac{1}{2}[1-\cos 2(A+B)] $ $=\frac{1}{2}\left[ 1-\frac{1-{{\tan }^{2}}(A+B)}{1+{{\tan }^{2}}(A+B)} \right] $ $=\frac{1}{2}\left[ \frac{1+{{\tan }^{2}}(A+B)-{{\tan }^{2}}(A+B)}{1+{{\tan }^{2}}(A+B)} \right] $ $=\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}}}} $ $=\frac{{{a}^{2}}}{{{a}^{2}}+{{(1-b)}^{2}}} $