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 \left.\sec ^{2} \theta-\tan ^{2} \theta=1\right] $
$\therefore 2 \sec \theta=2$
$\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.