# In a triangle ABC , angle C=90 degree , if tan (A/2) and tan (B/2) are the roots of the equation ax^2+bx+c=0 where a is not equal to 0 ,then the correct relationship is

Solution:    $A+B=90^{\circ}$    $\Rightarrow$       $\frac{1}{2}(A+B)=45^{\circ}$

Now ,      $\tan (\frac{A}{2})+\tan(\frac{B}{2})=-\frac{b}{a}$

$\tan (\frac{A}{2})\tan(\frac{B}{2})=\frac{c}{a}$

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

$1=\tan(\frac{A+B}{2})=\frac{-\frac{b}{a}}{1-\frac{c}{a}}\Rightarrow a-c=-b$

$\Rightarrow$                         $c=a+b$

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