If `sintheta and costheta " are the roots of the equation " mx^(2)+nx+1=0`, then find the relation between m and n.

Options

A. `n^(2)+m^(2)=4m`
B. `n^(2)-m^(2)=2m`
C. `n^(2)-m^(2)=5m`
D. `n^(2)+m^(3)=3m`

Answer

B

Explanation

The given equation is `mx^(2)+nx+1=0`
Here, a= m, b= n and c= 1
`sintheta+costheta =(-b)/(a)=(-n)/(m)`
`sintheta+costheta =(c)/(a)=(1)/(m)`
Consider,
`sintheta+costheta =(-n)/(m)`
or `(sintheta+costheta)^(2) =((-n)/(m))^(2)`
`sin^(2)theta+costheta^(2)+2sinthetacostheta =n^(2)/m^(2)`
`1+sinthetacostheta =n^(2)/m^(2) (therefore sin^(2)theta+cos^(2)theta=1)`
`1+2((1)/(m))=n^(2)/(m^(2)) (therefore sintheta costheta= (1)/(m))`
`1+(2)/(m)=n^(2)/(m^(2))`
`rArrn^(2)-m^(2)=2m`.