MCQ
In $\triangle ABC ,( a - b )^2 \cos ^2 \frac{ C }{2}+( a + b )^2 \sin ^2 \frac{ C }{2}=$
  • A
    $a^2$
  • B
    $b^2$
  • $c^2$
  • D
    bc

Answer

Correct option: C.
$c^2$
(C) $\left(a^2+b^2-2 a b\right) \cos ^2 \frac{C}{2}+\left(a^2+b^2+2 a b\right) \sin ^2 \frac{C}{2}$
$=\left(a^2+b^2\right)\left(\cos ^2 \frac{C}{2}+\sin ^2 \frac{C}{2}\right)$ $-2 a b\left(\cos ^2 \frac{C}{2}-\sin ^2 \frac{C}{2}\right)$
$=a^2+b^2-2 a b \cos C$
$=a^2+b^2-\left(a^2+b^2-c^2\right)=c^2$

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