Question

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

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

Answer 1

The correct option is A $1-2\sin A.\sin B.\cos C$

$A+B+C=\pi$

$A+B=\pi -C$

$\Rightarrow \cos (A+B)=\cos (\pi -c)$

$\Rightarrow \cos A\cos B-\sin A\sin B=-\cos C$

$\Rightarrow \left(\cos A\cos B\right)=(\sin A\sin B-\cos C)$

squaring lroth sides

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

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

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

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

$\Rightarrow 1-{\sin }^{2}A-{\sin }^{2}B+{\sin }^{2}A\cdot {\sin }^{2}B={\sin }^{2}A\cdot {\sin }^{2}B+{\cos }^{2}C$

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

$\Rightarrow 1+2\sin A\cdot \sin B\cdot \cos C={\sin }^{2}A+{\sin }^{2}B+{\cos }^{2}C$

$\Rightarrow 1+2\sin A\cdot \sin B\cdot \cos C=1-{\cos }^{2}A+1-{\cos }^{2}B+{\cos }^{2}C$

$\Rightarrow 2\sin A\sin B\cdot \cos C-1=-{\cos }^{2}A-{\cos }^{2}B+{\cos }^{2}C$

$\Rightarrow 1-2\sin A\cdot \sin B\cdot \cos C={\mathrm{Cos}}^{2}A+{\cos }^{2}B-{\cos }^{2}C$

: Answer: option (A)