Question

If $A+B+C={90}^{\circ }$, then ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=?$

• A. $1+4\sin A\sin B\sin C$
• B. $1-2\sin A\sin B\sin C$
• C. $2+2\sin A\sin B\sin C$
• D. $4\sin A\sin B\cos C$

Answer 1

Answer: (B)

$1-2\sin A\sin B\sin C$

Given

$A+B+C={90}^{\circ }$

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

$⟹1-{\cos }^{2}A+{\sin }^{2}B+{\sin }^{2}C$

$⟹1-({\cos }^{2}A-{\sin }^{2}B)+{\sin }^{2}C$

$⟹1-\cos (A+B)\cos (A-B)+{\sin }^{2}C$

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

$⟹1-\cos (90-C)\cos (A-B)+{\sin }^{2}C$

$⟹1-\sin \left(C\right)\cos (A-B)+{\sin }^{2}C$

$⟹1-\sin C\left(\cos \right(A-B)+\sin C)$

$⟹1-\sin C\left(\cos \right(A-B)+\sin (90-(A+B)\left)\right)$

$⟹1-\sin C\left(\cos \right(A-B)+\cos (A+B\left)\right)$

$(\cos C+\cos D=2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right))$

$⟹1-2\sin A\sin B\sin C$