Q68 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics

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

  • A
    $\cos \mathrm{B}$
  • B
    $\sin \mathrm{C}$
  • Correct answer
    C
    $\sin \mathrm{B}$
  • D
    $\cos C$

Solution

$(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2}=a^2+b^2$ $\begin{aligned} & \Rightarrow(a-b)^2 \frac{(1+\cos C)}{2}+(a+b)^2 \frac{(1-\cos C)}{2}=a^2+b^2 \\ & \Rightarrow a^2+b^2-2 a b \cos C=a^2+b^2 \\ & \Rightarrow 2 a b \cos C=0 \Rightarrow C=90^{\circ} \\ & \Rightarrow A+B=90^{\circ} \Rightarrow A=90^{\circ}-\mathbb{F} \\ & \Rightarrow \cos A=\cos \left(90^{\circ}-B\right)=\sin B .\end{aligned}$

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

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