Question

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

• A. $\sec \theta$
• B. $\tan \theta$
• C. $\sin \theta$
• D. $\cos \theta$

Answer 1

The correct option is A $\sec \theta$

$\sum (a-2b+c)=0$, so one root is $1$

Now $\sec \theta +\tan \theta =1$,

Also, ${\sec }^2\theta -{\tan }^2\theta =1$ ( from identity)

$\therefore \sec \theta -\tan \theta =1\Rightarrow \sec \theta =1$

So, one root is $\sec \theta$