삼각함수 식의 값

$\displaystyle \sin \theta + \cos \theta = \frac{1}{3}$에서 양변을 제곱하면, $\displaystyle \sin ^{2} \theta + 2 \sin \theta \cos \theta + \cos ^{2} \theta = \frac{1}{9}$ $\displaystyle \rightarrow 1 + 2 \sin \theta \cos \theta = \frac{1}{9}$ $\displaystyle \rightarrow \sin \theta \cos \theta = - \frac{4}{9}$ $\displaystyle \frac{1}{\cos \theta} \left( \tan \theta + \frac{1}{\tan ^{2} \theta} \right)$$\displaystyle = \frac{1}{\cos \theta} \left( \frac{\sin \theta}{\cos \theta} + \frac{\cos ^{2} \theta}{\sin ^{2} \theta} \right)$ $\displaystyle = \left( \frac{\sin \theta}{\cos ^{2} \theta} + \frac{\cos \theta}{\sin ^{2} \theta} \right)$$\displaystyle = \left( \frac{\sin ^{3} \theta + \cos ^{3} \theta}{\cos ^{2} \theta \sin ^{2} \theta} \right)$ $\displaystyle = \frac{( \sin \theta + \cos \theta ) ( \sin ^{2} \theta - \sin \theta \cos \theta + \cos ^{2} \theta )}{\sin ^{2} \theta \cos ^{2} \theta}$ $\displaystyle = \frac{\frac{1}{3} \left\{ 1 - \left( - \frac{4}{9} \right) \right\}}{\left( - \frac{4}{9} \right) ^{2}}$$\displaystyle = \frac{39}{16}$