Question

$\frac{{\sin }^{2}40°+{\sin }^{2}50°}{{\cos }^{2}20°+{\cos }^{2}70°}+\tan 10°\tan 20°\tan 60°\tan 70°\tan 80°$

Answer 1

$$\begin{gathered} \begin{array}{l}\frac{{\sin }^2{40}^0+{\sin }^2{50}^0}{{\cos }^2{20}^0+{\cos }^2{70}^0}+\tan {10}^0\tan {20}^0\tan {60}^0\tan {70}^0\tan {80}^0\\ \end{array} \\ \begin{array}{l}\text{=}\frac{{\sin }^2{40}^0+{\sin }^2({90}^0-{40}^0)}{{\cos }^2{20}^0+{\cos }^2({90}^0-{20}^0)}+\left(\tan {60}^0\right)\tan {20}^0\tan ({90}^0-{20}^0)\tan {10}^0\tan ({90}^0-{10}^0)\\ \end{array} \\ \begin{array}{l}\text{=}\frac{{\sin }^2{40}^0+{\cos }^2{40}^0}{{\cos }^2{20}^0+{\sin }^2{20}^0}+\sqrt{3}\left(\tan {20}^0\cot {20}^0\tan {10}^0\cot {10}^0\right)\\ \end{array} \\ \begin{array}{l}\text{=1+}\sqrt{3}\left(\tan {20}^0\frac{1}{\tan {20}^0}\tan {10}^0\frac{1}{\tan {10}^0}\right)\\ \end{array} \\ \text{=}\left(\text{1+}\sqrt{3}\right) \end{gathered}$$