# Eliminate $\theta$ from the equation $\sec \theta + \tan \theta = a+\sqrt b,$

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.

• Arent you given one more equation? Jan 25, 2021 at 16:12
• No sir, there is only one equation and no other equations are given. Jan 25, 2021 at 16:19
• @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. Jan 25, 2021 at 17:06

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

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

• 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? Jan 25, 2021 at 17:54
• 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. Jan 25, 2021 at 18:17
• Thank you! This is great Jan 25, 2021 at 18:20