Question

if sin(pie/4cotθ)=cos(pie/4tanθ) then θ

Answer 1

The solutions are given by the set of all $\theta$ such that $\sin(2\theta) = \frac{1}{4n+1}$ or $\cot(2\theta) = 4n+1$, for any integer $n$.

Answer 2

The equation $\sin A = \cos B$ is equivalent to $\sin A = \sin(\frac{\pi}{2}-B)$. This leads to $A = \frac{\pi}{2}-B + 2n\pi$ or $A = \pi - (\frac{\pi}{2}-B) + 2n\pi$. Substituting $A = \frac{\pi}{4}\cot\theta$ and $B = \frac{\pi}{4}\tan\theta$ and simplifying, we obtain two general forms for the solutions: $\sin(2\theta) = \frac{1}{4n+1}$ or $\cot(2\theta) = 4n+1$, where $n$ is an integer.