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) \[\frac{{{b}^{2}}}{{{a}^{2}}+{{(1-b)}^{2}}}\]
E) \[\frac{{{a}^{2}}}{{{b}^{2}}+{{(1-a)}^{2}}}\]
Correct Answer: A
Solution :
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, \[{{\sin }^{2}}(A+B)=\frac{1}{2}[1-\cos 2(A+B)]\] \[=\frac{1}{2}\left[ 1-\frac{1-{{\tan }^{2}}(A+B)}{1+{{\tan }^{2}}(A+B)} \right]\] \[=\frac{1}{2}\left[ \frac{1+{{\tan }^{2}}(A+B)-{{\tan }^{2}}(A+B)}{1+{{\tan }^{2}}(A+B)} \right]\] \[=\left[ \frac{{{\tan }^{2}}(A+B)}{1+{{\tan }^{2}}(A+B)} \right]=\frac{{{a}^{2}}/{{(1-b)}^{2}}}{\frac{[{{a}^{2}}+{{(1-b)}^{2}}]}{{{(1-b)}^{2}}}}\] \[=\frac{{{a}^{2}}}{{{a}^{2}}+{{(1-b)}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
