Question

Calculate $\frac{\text{sin}{70}^{\circ }}{\text{cos}{20}^{\circ }}+\frac{\text{cosec}{36}^{\circ }}{\text{sec}{54}^{\circ }}-2\text{cos}{43}^{\circ }\text{cosec}{47}^{\circ }$.

Answer 1

$\frac{\text{sin}{70}^{\circ }}{\text{cos}{20}^{\circ }}+\frac{\text{cosec}{36}^{\circ }}{\text{sec}{54}^{\circ }}-2\text{cos}{43}^{\circ }\text{cosec}{47}^{\circ }$

$=\frac{\text{sin}{70}^{\circ }}{\text{cos}({90}^{\circ }-{70}^{\circ })}+\frac{\text{cosec}{36}^{\circ }}{\text{sec}({90}^{\circ }-{36}^{\circ })}-2\text{cos}{43}^{\circ }\text{cosec}({90}^{\circ }-{43}^{\circ })$

$=\frac{\text{sin}{70}^{\circ }}{\text{sin}{70}^{\circ }}+\frac{\text{cosec}{36}^{\circ }}{\text{cosec}{36}^{\circ }}-2\text{cos}{43}^{\circ }\text{sec}{43}^{\circ }$

$\because cos({90}^{\circ }-\theta )=sin\theta ,sec({90}^{\circ }-\theta )=cosec\theta ,cosec({90}^{\circ }-\theta )=sec\theta )$

$=1+1-2\text{cos}{43}^{\circ }\text{sec}{43}^{\circ }$

$=1+1-2(\because \text{cos}\theta \text{sec}\theta =1)$

$=0$