Question

Without using tables, evaluate:
(i) sin ${53}^{\circ }÷cos{37}^{\circ }$
(ii) cos ${49}^{\circ }÷sin{41}^{\circ }$
(iii) tan ${66}^{\circ }÷cot{24}^{\circ }$

Answer 1

(i) $\frac{sin{53}^{\circ }}{cos{37}^{\circ }}=\frac{sin({90}^{\circ }-{37}^{\circ })}{cos{37}^{\circ }}=\frac{cos{37}^{\circ }}{cos{37}^{\circ }}=1$ [ $\because sin({90}^{\circ }-\theta )=cos\theta$ ]

(ii) $\frac{cos{49}^{\circ }}{sin{41}^{\circ }}=\frac{cos({90}^{\circ }-{41}^{\circ })}{sin{41}^{\circ }}=\frac{sin{41}^{\circ }}{sin{41}^{\circ }}=1$ [ $\because cos({90}^{\circ }-\theta )=sin\theta$ ]

(iii) $\frac{tan{66}^{\circ }}{cot{24}^{\circ }}=\frac{tan({90}^{\circ }-{24}^{\circ })}{cot{24}^{\circ }}=\frac{cot{24}^{\circ }}{cot{24}^{\circ }}=1$ [ $\because tan({90}^{\circ }-\theta )=cot\theta$ ]