If sin a and cos a are the roots of px^2 + qx + r=0 then..which is correct

p^2 - q^2 +2pr=0 p^2 + q^2 - 2pr=0

Question

If sin a  and cos a are the roots of px^2 + qx + r=0 then..which is correct

p^2 - q^2 +2pr=0      p^2 + q^2 - 2pr=0

anonymous · Answer

now can you do it???

anonymous · Answer

i think i can plzz  wait

anonymous · Answer

@THE_PROPHET i thinku have written it wrong

anonymous · Answer

kkkk....

anonymous · Answer

i get it sorry!!

anonymous · Answer

Can u correct it!

anonymous · Answer

$$(\sin \theta + \cos \theta)^{^{2}} = (\alpha + \beta)^{2}$$

anonymous · Answer

first substitute sina in place of x.....get an equation....
the substitute cos a ...get an equation......add both of them.....
psin2a+qsina+x=0

pcos2a+qcosa+r=0

anonymous · Answer

$$\sin \theta X \cos \theta = r/p$$

anonymous · Answer

adding both of them...
P+q(sina +cosa)+2r=0

but sina+cos a =-q/p(sum of roots)

p+q(-q/p)+2r=0
p^2-q^2+2rp=o

anonymous · Answer

@THE_PROPHET  i cant understand

anonymous · Answer

because sina and cosa ARE ROOTS OF THE EQUATION....they satisfy the equtaion....

so we can substitute sina and cosa in place of x to get the same value of 0....

|dw:1340984410787:dw|