Question

$a^2\cos 2B+b^2\cos 2A+2ab\cos (A-B)=c^2$
Is this statement true or false ?

• A. True
• B. False

Answer 1

The correct option is A True

True.
$a^2({\cos }^{2}B-{\sin }^{2}B)+b^2({\cos }^{2}A-{\sin }^{2}A)+2ab(\cos A\cos B+\sin A\sin B)$$=(a^2{\cos }^{2}B+b^2{\cos }^{2}A+2ab\cos A\cos B)-(a^2-{\sin }^{2}B+b^2{\sin }^{2}A-2ab\sin A\sin B)$
$={\left(a\cos B+b\cos A\right)}^2-{(a\sin B-b\sin A)}^2=c^2+0$
by projection formula and sine rule.