Question

If $A+B+C=\pi ,$ then $1+{\cos }^{2}C-{\cos }^{2}A-{\cos }^{2}B$ is equal to

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

Answer 1

Answer: (D)

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

Given

$A+B+C=\pi$

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

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

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

We know that

$\sin (x+y).\sin (x-y)={\sin }^{2}x-{\sin }^{2}y={\cos }^{2}y-{\cos }^{2}x$

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

$={\sin }^2(\pi -(A+C\left)\right)+\sin (A+C).\sin (A-C)$

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

$=\sin (A+C)\left[\sin \right(A+C)+\sin (A-C\left)\right]$

$=\sin (A+C)[2\sin \left(\frac{2A}{2}\right)+\cos \left(\frac{2C}{2}\right)]$

$=\sin (A+C)[2.\sin A.\cos (C\left)\right]$

$=\sin (\pi -B).2.\sin A.\cos C$

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