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. $\tan \theta$
• B. $\sec \theta$
• C. $-\sec \theta$
• D. $\sin \theta$

Answer 1

Answer: (B)

Given, $\sec \theta +\tan \theta =1$
$\Rightarrow \sec \theta -\tan \theta =1$
$[\because {\sec }^2\theta -{\tan }^2\theta =1]$
$\therefore 2\sec \theta =2$
$\Rightarrow \sec \theta =1$
$\therefore \cos \theta =1$
Clearly, $1$ is a root of given quadratic equation.
$\therefore \sec \theta$ and $\cos \theta$ are roots of given quadratic equation.