Q66 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics

In $\triangle \mathrm{ABC}, a^2 \sin 2 B+b^2 \sin 2 A=$

  • A
    $2 a b \cos \mathrm{~A}$
  • B
    $2 a b \sin \mathrm{~A}$
  • Correct answer
    C
    $2 a b \sin \mathrm{C}$
  • D
    $2 a b \cos \mathrm{C}$

Solution

We know that $\begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=K \\ & \Rightarrow a=K \sin A, b=K \sin B, c=K \sin C \\ & \therefore a^2 \sin 2 B+b^2 \sin 2 A \\ & =K^2\left[\sin ^2 A \cdot 2 \sin B \cdot \cos B+\sin ^2 B \cdot 2 \cdot \sin A \cdot \cos A\right] \\ & =2 K^2 \sin A \cdot \sin B[\sin A \cdot \cos B+\cos A \cdot \sin B] \\ & =2 a b \sin (A+B)=2 a b \sin (\pi-C)=2 a b \sin C\end{aligned}$

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

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