Question

If $x+y=\pi +z$, then $si{n}^{2}x+si{n}^{2}y-si{n}^{2}z=$

• A. 2 sin x sin y sin z
• B. 2 sin x sin y cos z
• C. 2 sin x cos y cos z
• D. 2 sin x cos y sin z

Answer 1

The correct option is B

2 sin x sin y cos z

$si{n}^{2}x+si{n}^{2}y-si{n}^{2}z,x+y=\pi -z=si{n}^{2}x+sin(y+z)sin(y-z)=si{n}^{2}x+sin(y+z)sin(\pi -x)=si{n}^{2}x+sin(y+z)sinx=sinx(sinx+sin(y+z\left)\right)=sinx\left(sin\right(\pi +z-4)+sin(y+z)$
$=sinx\left(sin\right(\pi +z-4)+sin(y+z\left)\right)$
=sinx (-sin (z-4)+sin(y+z))
=sin (2 cos z sin y)