Question

If `x+y=z`, then `cos^(2)x+cos^(2)y+cos^(2)z-2cosx.cosy.cosz` is equal to
A. `cos^(2)z`
B. `sin^(2)z`
C. 0
D. 1

Comments

z=x+y


cos2x+cos2y+cos2z−2cosx.cosy.cosz


=cos2x+cos2y+cos2(x+y)−2cosx.cosy.cos(x+y)


=cos2x+cos2y+(cosxcosy−sinxsiny)2−2cosxcosy[cosxcosy−sinxsiny]


=cos2x+cos2y+cos2xcos2y+sin2xsin2y−2cosxcosysinxsiny−2cos2xcos2y+2cosxcosysinxsiny


=cos2x+cos2y−cos2xcos2y+sin2xsin2y


=cos2x+cos2y−cos2xcos2y+sin2x(1−cos2y)


=cos2x+cos2y−cos2xcos2y+sin2x−sin2xcos2y


=cos2x+sin2x+cos2y−cos2y(cos2x+sin2x)


=1

Answer 1

Correct Answer - D
`cos^(2)x+cos^(2)y+cos^(2)z-cosz(cos(x+y)+cos(x-y))`
`=cos^(2)x+cos^(2)y+cos^(2)z-cos^(2)z-cos(x+y).cos(x-y)=1`

Comments

z=x+y


cos2x+cos2y+cos2z−2cosx.cosy.cosz


=cos2x+cos2y+cos2(x+y)−2cosx.cosy.cos(x+y)


=cos2x+cos2y+(cosxcosy−sinxsiny)2−2cosxcosy[cosxcosy−sinxsiny]


=cos2x+cos2y+cos2xcos2y+sin2xsin2y−2cosxcosysinxsiny−2cos2xcos2y+2cosxcosysinxsiny


=cos2x+cos2y−cos2xcos2y+sin2xsin2y


=cos2x+cos2y−cos2xcos2y+sin2x(1−cos2y)


=cos2x+cos2y−cos2xcos2y+sin2x−sin2xcos2y


=cos2x+sin2x+cos2y−cos2y(cos2x+sin2x)


=1