Question

In $△ABC,b^{2}cos2A-a^{2}cos2B$ =

• A. $b^2-a^2$
• B. $b^2-c^2$
• C. $c^2-a^2$
• D. $a^2+b^2+c^2$

Answer 1

The correct option is A $b^2-a^2$

$b^{2}cos2A-a^{2}cos2B=b^2(1-2si{n}^{2}A)-a^2(1-2si{n}^{2}B)$
$=b^2-a^2-2(b^{2}si{n}^{2}A-a^{2}si{n}^{2}B)=b^2-a^2.$