What is the value of \(\frac{\sin(18^\circ)\cos(72^\circ)+\sec(61^\circ)\sin(29^\circ)+\cos(18^\circ)\sin(72^\circ)+\operatorname{cosec}(61^\circ)\cos(29^\circ)}{\sec^2(48^\circ)-\cot^2(42^\circ)}\)?

Solution

Solution: Using \(\sin A\cos B+\cos A\sin B=\sin(A+B)\), the first and third terms sum to \(\sin90^\circ=1\). Also, \(\sec61^\circ\sin29^\circ=1\), because \(\cos61^\circ=\sin29^\circ\), and \(\operatorname{cosec}61^\circ\cos29^\circ=1\), because \(\sin61^\circ=\cos29^\circ\). The numerator is 3. The denominator is \(\sec^2 48^\circ-\tan^2 48^\circ=1\), since \(\cot42^\circ=\tan48^\circ\). Therefore, the value is 3. Trick: Use complementary-angle identities and \(\sec^2 A-\tan^2 A=1\).