Question

If $sec\theta +tan\theta =1$ then one root of equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ is

• A. tanθ
• B. secθ
• C. sinθ
• D. none of these

Answer 1

The correct option is B secθ

$sec\theta +tan\theta =\mathrm{1.......}\left(1\right)$
We have that
$se{c}^2\theta -ta{n}^2\theta =1$
$sec\theta -tan\theta =1........\left(2\right)$
from (1) and (2)
$Sec\theta =1$
$a(b-c)x^2+b(c-a)x+c(a-b)=0$
$abx(x-1)-ac(x^2-1)+bc(x-1)=0$
$x=1=sec\theta$