Question

Evaluate $\frac{\cos {65}^{\circ }\sin {18}^{\circ }\cos {58}^{\circ }}{\cos {72}^{\circ }\sin {25}^{\circ }\sin {32}^{\circ }}$

Answer 1

Using trigonometric ratios of complementary angles, we get
$\cos {65}^{\circ }=\cos ({90}^{\circ }-{25}^{\circ })=\sin {25}^{\circ }$,
$\sin {18}^{\circ }=\sin ({90}^{\circ }-{72}^{\circ })=\cos {72}^{\circ }$
$\cos {58}^{\circ }=\cos ({90}^{\circ }-{32}^{\circ })=\sin {32}^{\circ }$.
$\therefore \frac{\cos {65}^{\circ }\sin {18}^{\circ }\cos {58}^{\circ }}{\cos {72}^{\circ }\sin {25}^{\circ }\sin {32}^{\circ }}=\frac{\sin {25}^{\circ }\cos {72}^{\circ }\sin {32}^{\circ }}{\cos {72}^{\circ }\sin {25}^{\circ }\sin {32}^{\circ }}=1$