Voltage across inductor in AC
The voltage across an inductor is V = \$L\dfrac{di}{dt}\$ so if the steady state voltage is a sinewave then the current, when differentiated is a sine wave. This means the current is an inverted cosine wave as shown in the picture.


  The voltage across inductor is V=IXL


If you are reffering to RMS voltage and current then that is correct but, of course RMS voltages and currents do not embody the notion of phase angle (as seen in the diagram above). So the ratio of the voltage and current magnitudes equals the magnitude of the impedance but, behind this simplistic notion lies a deeper truth.