Question

If $\sec \theta +\tan \theta =1$, then a 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

Answer: (A), (D)

$\sec \theta$
$\cos \theta$

Let us consider the equation, $(a-2b+c)x^2+(b-2c+a)x+(c-2a+b)=0$.
We can observe that, $x=1$ is a root of the equation. While the other root depends on the values of $a,b,c$.....(1)

Let us consider, $sec\theta +\tan \theta =1$
$\Rightarrow 1+\sin \theta =\cos \theta$ ( $\cos \theta \ne 0$)
On squaring we get,
$1+2\sin \theta +{\sin }^2\theta =1-{\sin }^2\theta$
$\Rightarrow {\sin }^2\theta +\sin \theta =0$
$\Rightarrow \sin \theta =-1$ or $\sin \theta =0$
We will reject $\sin \theta =-1$ as $\cos \theta$ can not be zero.
For $\sin \theta =0$ , $\cos \theta =1$ (Substituting in the given trigonometric equation).......(2)
Hence from (1) and (2), options A and D would be a root of the given quadratic equation.