Question

If $(tan^{-1} x)^2 + (cot^{-1}x)^2 = \frac{5x\pi^2}{8}$, x=?

Answer 1

Given that $tan^{-1}(x) + cot^{-1}(x) = \frac{2π}{2}$, the equation can be rearranged as:

$(tan^{-1}(x) + cot^{-1}(x))^2 - 2tan^{-1}(x)(\frac{2π}{2} - tan^{-1}(x)) = \frac{8}{(\frac{5π}{2})}$

Simplifying further: $2(tan^{-1}(x))^2 - 2(\frac{2π}{2})tan^{-1}(x) - \frac{8}{(\frac{3π}{2})} = 0$

This can be rewritten as: $2(tan^{-1}(x))^2 - 2(π)tan^{-1}(x) - \frac{8}{(\frac{3π}{2})} = 0$

From this equation, we can deduce that $tan^{-1}(x) = \frac{-4π}{3}$ or $\frac{4π}{3}$.

Hence, the solutions are $x = -1$.