Why does cos⁡θ=sin⁡(π2−θ) even when θ>90∘?I get why this relationship holds true in 90 degree...

Draw a centre-O circle of radius 1 and add in the radius with an end at (1,0), and rotate that radius through an angle ${\displaystyle \theta }$ anticlockwise so its end becomes ${\displaystyle (\cos \theta ,\sin \theta )}$. You could have got the same result by rotating the radius at (0,1) clockwise through ${\displaystyle \frac{\pi }{2}-\theta }$, so ${\displaystyle (\cos \theta ,\sin \theta )=(\sin (\frac{\pi }{2}-\theta ),\cos (\frac{\pi }{2}-\theta ))}$

Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive
${\displaystyle \cos x=\sin (\frac{\pi }{2}-x)}$
for ${\displaystyle 0<x<\frac{\pi }{2}}$.
Then one would learn the Maclaurin series of ${\displaystyle \sin }$ and ${\displaystyle \cos }$, and then one can show that they have an infinite radius of convergence.
Since ${\displaystyle (0,\frac{\pi }{2})\subset \mathbb{C}}$ and ${\displaystyle \cos x=\sin (\frac{\pi }{2}-x)}$ in this interval, by identity theorem this immediately implies the two functions ${\displaystyle \left(\cos x\text{ and }\sin (\frac{\pi }{2}-x)\right)}$ are equal for any ${\displaystyle x\in \mathbb{C}}$