# Thread: Prove that trigonometric terms are in arithmetic progression?

1. ## Prove that trigonometric terms are in arithmetic progression?

If $\displaystyle \theta_1$ and $\displaystyle \alpha$ are two real numbers such that $\displaystyle \frac{\cot^4 \theta}{\csc^4 \theta}\sec^2 \alpha$, $\displaystyle \frac{1}{\csc 150^{\circ}}$, $\displaystyle \frac{\tan^4 \theta}{\sec^4 \theta}\csc^2 \alpha$ are in Arithmetic Progression, prove that:

$\displaystyle \frac{\sec^{2n} \alpha}{\sec^{2n + 2} \theta}$, $\displaystyle \frac{1}{\csc 150^{\circ}}$, $\displaystyle \frac{\csc^{2n} \alpha}{\csc^{2n + 2} \theta}$ are also in A.P. for all $\displaystyle n\in \mathbb{N}$

2. ## Vacuously True?

By definition, a,b,c are in arithmetic progression iff c-b=b-a ~ a+c=2b.

Given $\displaystyle \frac{\cot^4 \theta}{\csc^4 \theta}\sec^2 \alpha$ + $\displaystyle \frac{\tan^4 \theta}{\sec^4 \theta}\csc^2 \alpha$ = $\displaystyle \frac{2}{\csc 150^{\circ}}$

Or, $\displaystyle \frac{\sec^2 \alpha}{\sec^4\theta}$ + $\displaystyle \frac{\csc^2 \alpha}{\csc^4\theta}$ = $\displaystyle -\frac{4\sqrt{3}}{3}$

But the sum of two squares can never equal a negative number, therefore there are no real solutions for $\displaystyle \theta$ and $\displaystyle \alpha$. Ergo, this theorem is vacuously true.