# Explain this equality: &#x2212;<!-- - --> 90 &#x2218;<!-- ∘ --> </msup> +

Explain this equality: $-{90}^{\circ }+{\mathrm{tan}}^{-1}\frac{{X}_{C}}{R}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}-{\mathrm{tan}}^{-1}\frac{R}{{X}_{C}}$
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$\frac{1}{x}=\mathrm{tan}\theta$
$x=\mathrm{cot}\theta =\mathrm{tan}\left(\frac{\pi }{2}-\theta \right)$
$\mathrm{arctan}x=\frac{\pi }{2}-\theta$
$\mathrm{arctan}x=\frac{\pi }{2}-\theta$
So,
$\mathrm{arctan}x=\frac{\pi }{2}-\mathrm{arctan}\frac{1}{x}$