Q67 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics

In $\triangle A B C$, $(a-b)^2 \sin ^2\left(\frac{A+B}{2}\right)+(a+b)^2 \sin ^2\left(\frac{C}{2}\right)=$

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

Solution

$\begin{aligned} & \text { In } \triangle A B C, A+B+C=\pi \\ & \qquad A+B=\pi-C \\ & =(a-b)^2 \sin ^2\left(\frac{A+B}{2}\right)+(a+b)^2 \sin ^2 \frac{C}{2} \\ & =\left[a^2+b^2-2 a b\right] \sin ^2\left(\frac{\pi-C}{2}\right) \\ & +\left[a^2+b^2+2 a b\right] \sin ^2 \frac{C}{2}\end{aligned}$ $\begin{aligned} & =\left(a^2+b^2\right) \cos ^2 \frac{C}{2}-2 a b \cos ^2 \frac{C}{2} \\ & +\left(a^2+b^2\right) \sin ^2 \frac{C}{2}+2 a b \sin ^2 \frac{C}{2} \\ & =a^2+b^2-2 a b\left(\cos ^2 \frac{C}{2}-\sin ^2 \frac{C}{2}\right) \\ & =a^2+b^2-2 a b \cos C \\ & =a^2+b^2-2 a b\left(\frac{a^2+b^2-c^2}{2 a b}\right) \\ & \Rightarrow a^2+b^2-a^2-b^2+c^2=c^2\end{aligned}$

Hence, the correct answer is option: 'C'.

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