Q149 / 200 · TS EAMCET 2001 · Mathematics
In a $\triangle A B C, a^2 \sin 2 C+c^2 \sin 2 A$ is equal to
- A$\Delta$
- B$2 \Delta$
- C$3 \Delta$
- Correct answerD$4 \Delta$
Solution
$\begin{aligned} & a^2 \sin 2 C+c^2 \sin 2 A \\ & =4 R^2\left[\sin ^2 A-2 \sin C \cos C\right.\end{aligned}$
$\begin{aligned} & \left.+\sin ^2 C \cdot 2 \sin A+\cos A\right] \\ & \quad=4 R^2 \cdot 2 \sin A \sin C[\cos C \sin A+\sin C \cos A] \\ & \quad=8 R^2 \sin A \sin C \sin (A+C)\end{aligned}$
$\begin{aligned} & =8 R^2 \sin A \sin B \sin C \\ & =8 R^2 \cdot \frac{a}{2 R} \cdot \frac{b}{2 R} \cdot \frac{c}{2 R} \\ & =\frac{a b c}{R}=4 \Delta\end{aligned}$
Hence, the correct answer is option: 'D'.