Question

If $A+B+C={180}^o$, then ${\sin }^{2}A+{\sin }^{2}B-{\sin }^{2}C=?$

• A. $2\cos A\sin B\sin C$
• B. $2\cos A\sin B\sin C$
• C. $2\sin A\cos B\sin C$
• D. $2\sin A\sin B\cos C$

Answer 1

The correct option is A $2\cos A\sin B\sin C$

$\begin{array}{cc}& \text{Given that,}A+B+C=180\\ & \therefore \sin 2A+\sin 2B-\sin 2C\\ \Rightarrow & \frac{1-\cos 2A}{2}+\frac{1-\sin 2B}{2}-\frac{1-\cos 2C}{2}\\ \Rightarrow & \frac{1}{2}[1-(\cos 2A+\cos 2B-\cos 2C\left)\right]\\ \Rightarrow & \frac{1}{2}[1-\{\left(2\cos \right(A+B)\cdot \cos (A-B)-2\cos 2C+1)\left\}\right]\\ \Rightarrow & \frac{1}{2}[2\cos C\cdot \cos (A-B)+2\cos 2C]\\ \Rightarrow & \frac{1}{2}\left[2\cos C\right(\cos (A-B)+\cos C]\\ \Rightarrow & \frac{1}{2}\left[2\cos C\right(\cos (A-B)-\cos (A+B)\left)\right]\\ \Rightarrow & \frac{1}{2}\left[2\cos C\right[-2\sin A\cdot \sin (-B)\left]\right]\\ \Rightarrow & 2\sin A\sin B\sin C\\ & \text{hence,}\left(A\right)\text{is the correct option}\end{array}$