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.
Best Answer
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
$$b = a^{2}-1$$