If tan(x)=2ba-c,(a≠c),y=acos2(x)+2bsin(x)cos(x)+csin2(x) and z=asin2(x)-2bsin(x)cos(x)+ccos2(x),then
y=z
y+z=a+c
y-z=a-c
y-z=(a-c)2+4b2
Explanation for correct option
Given: tan(x)=2ba-c,(a≠c),y=acos2(x)+2bsin(x)cos(x)+csin2(x) and z=asin2(x)-2bsin(x)cos(x)+ccos2(x)
y=acos2(x)+2bsin(x)cos(x)+csin2(x)
z=asin2(x)-2bsin(x)cos(x)+ccos2(x)
∴y+z=acos2(x)+2bsin(x)cos(x)+csin2(x)+asin2(x)-2bsin(x)cos(x)+ccos2(x)⇒y+z=a(sin2(x)+cos2(x))+2bsin(x)cos(x)-2bsin(x)cos(x)+c(sin2(x)+cos2(x))⇒y+z=a+c∵sin2(x)+cos2(x)=1
Hence, option B is correct.
If ax=by=cz and b2=ac,then show that y=2zxz+x.