Question

Prove that sin 48° sec 42° + cos 48° cosec 42° = 2

Answer 1

We will simplify the left-hand side

${\displaystyle \sin 48°\sec 42°+\cos 48°\cos ec42°={\sin 48}^{\circ }.\sec ({90}^{\circ }-{48}^{\circ })+{\cos 48}^{\circ }\cos ec({90}^{\circ }-{48}^{\circ })}$

${\displaystyle ={\sin 48}^{\circ }.{\sec 48}^{\circ }+{\cos 48}^{\circ }.\cos ec{42}^{\circ }={\sin 48}^{\circ }.\sec ({90}^{\circ }-{48}^{\circ })+{\cos 48}^{\circ }.\cos ec({90}^{\circ }-{48}^{\circ })}$

${\displaystyle ={\sin 48}^{\circ }.{\cos 48}^{\circ }+{\cos 48}^{\circ }.{\sin 48}^{\circ }}$

= 1 + 1

= 2

Proved