A basic doubt while taking square roots of variables which have ± sign before the square root sign.

Question

Normally, we know that √x² = |x| as x is a variable here.

But if it is ± √x² , then we simply write it as ±x , because ± has specifically asked for 2 values. So on a higher level - it doesn't matter if mod is skipped. ( Correct me if I am going wrong )

Ok so the original question was .....

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Let −π/6 < θ < −π/12

Suppose α₁ & β₁ are the roots equation x² − 2x sec(θ) + 1 = 0

and α₂ & β₂ are roots of the equation x² + 2x tan(θ) − 1 = 0.

If α₁ > β₁ & α₂ > β₂, then find the value of α₁+β₂

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On Calculating the Roots of first equation we get them as

sec(θ) ± | tan(θ) |

and on Calculating the Roots of second equation we get them as

-tan(θ) ± | sec(θ) |

The given answer to this question is -2 tan(θ)

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Now I don't have any problem with α₂ & β₂, as one can clearly make out that α₂ is larger than β₂ as sec(θ) is anyhow positive in the 4th quadrant

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But I have problem with assigning α₁ & β₁ because √tan²(θ) is |tan θ| and I guess, in this specific case we can't ignore the mod "| |" sign, else α₁ & β₁ will interchange

If mod sign isn't ignored, then the answer of α₁+β₂ will be 0, as

sec(θ) + | tan(θ) | = α₁ - larger root

sec(θ) - | tan(θ) | = β₁ - smaller root

as it is given α₁ > β₁

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and if mod sign is ignored- which is what is done ( possibly due to ± sign in formula) then

sec(θ) + tan(θ) = β₁ - smaller root

sec(θ) - tan(θ) = α₁ - bigger root

as tan(θ) is negative in 4th quadrant the answer of α₁+β₂ comes to -2tan(θ) - which is given as the solution.

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So what do you think is the correct solution for the question ? Or is there anything I am specifically missing here, and so the mod was discarded ?

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This question was also asked in JEE Advanced ( India ) in 2016.

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Answer 1

In your section where you come up with $\alpha_1+\beta_2=0$, you have:

$\alpha_1 = \sec(\theta)+\mid\tan(\theta)\mid$ and

$\beta_1 = \sec(\theta)-\mid\tan(\theta)\mid$.

Also we then have $\alpha_2=\sec(\theta)+\mid \tan(\theta)\mid$

and $\beta_2 =-\sec(\theta)+\mid \tan(\theta)\mid$.

So $\alpha_1+\beta_2=2\mid \tan(\theta)\mid = -2\tan(\theta)$

(last equality because, as you noted, in the fourth quadrant, tangent takes on negative values).