If $\sec ^2 \alpha+4 \cos ^2 \alpha=4$ and $0^{\circ} \leq \alpha \leq 90^{\circ}$, then find the value of $\alpha$.

30° 0° 45° 60°

45°

4cos2α + \(\frac{1}{cos2α}\)  = 4 4cos4α + 1  = 4 cos2α 4cos4α - 4 cos2α + 1 = 0 4 cos2α ( cos2α - 1 ) -1 ( cos2α - 1 ) = 0 Now  cos2α - 1 = 0 and 4cos2α - 1 = 0 cos2α - 1 = 0 α = 90º  ( not possible ) 4cos2α - 1 = 0 cos2α = \(\frac{1}{4}\) So , α = 45º