1.

The value of `sec 40^(@)+sec 80^(@)+sec 160^(@)` will beA. 4B. `-4`C. 6D. 8

Answer» Correct Answer - C
We find that
`cos40^(@)+cos80^(@)cos160^(@)`
`2cos60^(@)cos20^(@)-cos20^(@)=0`
`cos40^(@)cos80^(@)+cos80^(@)cos160^(@)+160^(@)cos40^(@)`
`=1/2[2cos80^(@)cos40^(@)+2cos160^(@)cos80^(@)+2cos160^(@)cos40^(@)]`
`=1/2[cos120^(@)+cos40^(@)+cos240^(@)+cos80^(@)+cos200^(@)+cos120^(@)]`
`=1/2(-1/2+2cos40^(@)-1/2+cos80^(@)-cos20^(@)""1/2)`
`=1/2(-(3)/(2)+cos60^(@)cos20^(@)-cos20^(@))=-3/4` and, `cos40^(@)cos80^(@)cos160^(@)=(sin(2^(3)xx40^(@)))/(2^(3)sin40^(@))=1/8(sin320^(@))/(sin40^(@))=-1/8`
So, the equation having `cos40^(@),cos80^(@)and cos160^(@)` at its roots is
`x^(3)-x^(2)xx0+x xx-3/4-(-(1)/(8))=0or,8x^(3)-6x +1=0`
The equation having `sec 40^(@), sec80^(@) and sec 160^(@)` as its roots is `x^(3)-6x^(2)+8=0`
`thereforesec40^(@)+sec80^(@)+sec160^(@)=6`


Discussion

No Comment Found

Related InterviewSolutions