Question

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

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

Answer 1

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

Given

$A+B+C=\pi$

$⟹C=\pi -(A+B)$

${\sin }^{2}A+{\sin }^{2}B-{\sin }^{2}C$

${\sin }^{2}A+\sin (B+C)\sin (B-C)$

${\sin }^{2}A+\sin A\sin (B-C)$

$\sin A(\sin A+\sin (B-C\left)\right)$

$\sin A\left(\sin \right(B+C)+\sin (B-C\left)\right)$

$\sin A\left(2\sin B\cos C\right)$

$2\sin A\sin B\cos C$