Question

In $\Delta$ le $ABC$, if a,b and $A$ are given then there are two triangles are formed with third side $c_1$ and $c_2$ such that then the value of $(c_1-c_2{)}^2+(c_1+c_2{)}^2{\tan }^{2}A=$

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

Answer 1

The correct option is A $4a^2$

$cosA=\frac{c^2+b^2-a^2}{2bc}$
$c^2-2bccosA+b^2-a^2=0$
$c_1+c_2=2bcosA$
$c_1{c}_2=b^2-a^2$
$=(c_1+c_2{)}^2-4c_1{c}_2+(c_1+c_2{)}^{2}ta{n}^{2}A$
$=(c_1+c_2{)}^{2}cose{c}^{2}A-4c_1{c}_2$
$=(2bcosA{)}^{2}se{c}^{2}A-4(b^2-a^2)$
$=4b^2-4b^2+4a^2$
$=4a^2$