Question

In $\Delta ABC$, if $a,b$ and $A$ are given and there are two possibilities for the third side $c_1$ and $c_2,$ then the value of $(c_1-c_2{)}^2+(c_1+c_2{)}^2{\tan }^{2}A$ is equal to

• A. $a^2$
• B. $4a^2$
• C. $b^2$
• D. $4b^2$

Answer 1

The correct option is B $4a^2$

$a^2=b^2+c^2-2bc\cos A$
$\Rightarrow c^2-2b\cos A\cdot c+b^2-a^2=0$
Thus, above equation will be having roots $c_1,c_2$
Sum of roots
$\Rightarrow c_1+c_2=2b\cos A\cdots \left(i\right)$
Product of roots
$\Rightarrow c_1{c}_2=b^2-a^2\cdots \left(ii\right)$
Now, $(c_1-c_2{)}^2+(c_1+c_2{)}^2{\tan }^{2}A=(c_1+c_2{)}^2-4c_1{c}_2+(c_1+c_2{)}^2{\tan }^{2}A$
$\Rightarrow (c_1+c_2{)}^2[1+{\tan }^{2}A]-4(b^2-a^2)$
$\Rightarrow (c_1+c_2{)}^2[1+\frac{{\sin }^{2}A}{{\cos }^{2}A}]-4(b^2-a^2)$
$\Rightarrow 4b^2{\cos }^{2}A\left[\frac{1}{{\cos }^{2}A}\right]-4(b^2-a^2)$
$\Rightarrow 4b^2-4b^2+4a^2$
$\Rightarrow 4a^2$
$\therefore (c_1-c_2{)}^2+(c_1+c_2{)}^2{\tan }^{2}A=4a^2$