Question

If in triangle $ABC,\Delta =a^2-{\left(b-c\right)}^2$, then find the value of $tanA$.

Answer 1

Given,

$\Delta =a^2-(b-c{)}^2$

Now as we know that $s=\frac{a+b+c}{2}$

or, $a=2s-(b+c)$

putting this value of a in $\Delta =a^2-(b+c{)}^2$, we get

$\Delta =[2s-(b+c){]}^2-(b-c{)}^2$

$\Delta =[4s^2+(b+c{)}^2-4s(b+c)]-(b-c{)}^2$

$\Delta =4s^2-4s(b+c)+\left[\right(b+c{)}^2-(b-c{)}^2]$

$\Delta =4s^2-4s(b+c)+4bc$

$\Delta =4s^2-4sb-4sc+4bc$

$\Delta =4s(s-b)-4c(s-b)$

$\Delta =(4s-4c)(s-b)$

$\Delta =4(s-c)(s-b)$

$\frac{1}{4}=$ $\frac{(s-c)(s-b)}{\Delta }$ ...............1

Now as we know that $\tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$

$\sqrt{(s-b)(s-c)}=\tan \frac{A}{2}\sqrt{s(s-a)}$

multiply both side by $\sqrt{(s-b)(s-c)}$

$(s-b)(s-c)=$ $\tan \frac{A}{2}$ $\sqrt{s(s-a)(s-b)(s-c)}$

$(s-b)(s-c)=$ $\tan \frac{A}{2}$ $\Delta$

$\frac{(s-b)(s-c)}{\Delta }=$ $\tan \frac{A}{2}$ ................2

by using 1 and 2

$\frac{1}{4}=$ $\tan \frac{A}{2}$

$\tan A=\frac{2\tan \frac{A}{2}}{1-{\tan }^2\frac{A}{2}}$

$=\frac{2\left(\frac{1}{4}\right)}{1-{\left(\frac{1}{4}\right)}^2}$

$=\frac{8}{15}$