<s>[INST] Explain this equality: $-90^\circ + \tan^{-1}\frac{X_C}{R} = - \tan^{-1}\frac{R}{X_C}$ 
$$-90^\circ + \tan^{-1}\frac{X_C}{R} \;=\; - \tan^{-1}\frac{R}{X_C}$$

Please explain how both are equal.
 [/INST] $$\text{Let} \ \ \arctan \frac{1}{x} = \theta$$
$$\frac{1}{x} = \tan\theta$$
$$x = \cot \theta = \tan \left(\frac{\pi}{2} - \theta\right)$$
$$\arctan x = \frac{\pi}{2} - \theta$$
So, $$   \arctan x=\frac{\pi}{2}-\arctan\frac{1}{x}$$
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