Question #04552

May 12, 2017

Use $\sin \left(\theta\right) = \cos \left({90}^{\circ} - \theta\right)$. Answer: $2$

Explanation:

Evaluate ${\left(\sin {39}^{\circ} / \cos {51}^{\circ}\right)}^{2} + {\left(\cos {51}^{\circ} / \sin {39}^{\circ}\right)}^{2}$

Here, it is important to note that $\sin \left(\theta\right) = \cos \left({90}^{\circ} - \theta\right)$ and $\cos \left(\theta\right) = \sin \left({90}^{\circ} - \theta\right)$

Therefore, by substitution:

${\left(\sin {39}^{\circ} / \cos {51}^{\circ}\right)}^{2} + {\left(\cos {51}^{\circ} / \sin {39}^{\circ}\right)}^{2}$

$= {\left(\cos \frac{{90}^{\circ} - {39}^{\circ}}{\cos} {51}^{\circ}\right)}^{2} + {\left(\cos {51}^{\circ} / \cos \left({90}^{\circ} - {39}^{\circ}\right)\right)}^{2}$

$= {\left(\cos {51}^{\circ} / \cos {51}^{\circ}\right)}^{2} + {\left(\cos {51}^{\circ} / \cos {51}^{\circ}\right)}^{2}$

$= {1}^{2} + {1}^{2}$

$= 2$

Therefore, our answer is $2$.