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Eliminate $\theta$ from the equation $$\sec \theta + \tan \theta = a+\sqrt b,$$where $~a,b, \sec \theta \in \mathbb{Q},$ and $\sqrt b \notin \mathbb{Q}.$

My attempt: By the given condition we must have $~\tan \theta \notin \mathbb{Q}.$ Then we can have $$\tan \theta = x+y,~~ \text{ where} ~ x \in \mathbb{Q},~ y \notin \mathbb{Q}.$$ But after this I am not able to proceed suitably to eliminate $\theta$. I did square both side, took bar both side, but failed.

Please help me to solve this.

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    $\begingroup$ Arent you given one more equation? $\endgroup$ Jan 25, 2021 at 16:12
  • $\begingroup$ No sir, there is only one equation and no other equations are given. $\endgroup$
    – abcdmath
    Jan 25, 2021 at 16:19
  • $\begingroup$ @ultralegend5385 It is conceivable that restricting $a$, $b$, $\sec\theta$ to $\mathbb{Q}$ is enough to eliminate $\theta$ and play the role of a second equation. $\endgroup$
    – 2'5 9'2
    Jan 25, 2021 at 17:06

1 Answer 1

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You can use the identity $\tan(\theta) = \pm\sqrt{\sec^{2}(\theta)-1}$

Then you get $$\sec(\theta) + \sqrt{\sec^{2}(\theta)-1} = a + \sqrt{b}$$

Since $\sec(\theta)$ is in $Q$ and $b \in Q$ the negative version of the tangent identity is ruled out, then $\sec(\theta) = a$. So the equation reduces to

$$a + \sqrt{a^{2}-1} = a + \sqrt{b}$$

So

$$b = a^{2}-1$$

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  • $\begingroup$ Can you please give me a hint why $\sec \theta + \sqrt{\sec^2 \theta -1}=a+\sqrt b \implies \sec \theta =a$ ? Is $\sqrt{\sec^2 \theta -1}$ is necessarily irrational? $\endgroup$
    – abcdmath
    Jan 25, 2021 at 17:54
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    $\begingroup$ Ok. Subtract $sec(\theta)$ from both sides and then square both sides of the equation. $$sec^{2}(\theta) -1 = (a-sec(\theta))^{2} + b + 2(a-sec(\theta))\sqrt{b}$$ $$sec^{2}(\theta) -1 - (a-sec(\theta))^{2} - b = 2(a-sec(\theta))\sqrt{b}$$ If $a-sec(\theta)$ was non-zero rational then we could divide by it obtaining a contradiction that $\sqrt{b}$ is not rational. $\endgroup$ Jan 25, 2021 at 18:17
  • $\begingroup$ Thank you! This is great $\endgroup$
    – abcdmath
    Jan 25, 2021 at 18:20

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