If sec teta and tan teta are the roots of ax²+bx+c=0(a,b not equal to 0) ,then the value of sec teta- tan teta is

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If secθ+tanθ=1, then a root of the equation (a−2b+c)x

+(b−2c+a)x+(c−2a+b)=0 is

This question has multiple correct options

Hard

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Correct options are A) and D)

Let us consider the equation, (a−2b+c)x

+(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, secθ+tanθ=1

⇒1+sinθ=cosθ ( cosθ



=0)

On squaring we get,

1+2sinθ+sin

θ=1−sin

⇒sin

θ+sinθ=0

⇒sinθ=−1 or sinθ=0

We will reject sinθ=−1 as cosθ can not be zero.

For sinθ=0 , cosθ=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.