Question

$\frac{3\cos 55°}{7\sin 35°}-\frac{4(\cos 70°\cos ec20°)}{7(\tan 5°\tan 25°\tan 45°\tan 65°\tan 85°)}$

Answer 1

$$\begin{gathered} \begin{array}{l}\frac{3\cos {55}^0}{7\sin {35}^0}-\frac{4\left(\cos {70}^0\cos \text{ec}{20}^0\right)}{7\left(\tan 5^0\tan {25}^0\tan {45}^0\tan {65}^0\tan {85}^0\right)}\\ \end{array} \\ \begin{array}{l}=\frac{3\cos ({90}^0-{35}^0)}{7\sin {35}^0}-\frac{4\left(\cos {70}^0\frac{1}{\sin {20}^0}\right)}{7\times 1\{\tan 5^0\tan ({90}^0-5^0).\tan {25}^0\tan ({90}^0-{65}^0)\}}\\ \end{array} \\ \begin{array}{l}\text{=}\frac{3\sin 35}{7\sin {35}^0}-\frac{4\left(\cos {70}^0\frac{1}{\sin ({90}^0-{70}^0)}\right)}{7\left(\tan 5^0\cot 5^0\right)\left(\tan {25}^0\cot {25}^0\right)}\\ \end{array} \\ \begin{array}{l}\text{=}\frac{3}{7}-\frac{4\left(\cos {70}^0\frac{1}{\cos {70}^0}\right)}{7\left(\tan 5^0\frac{1}{\tan 5^0}\right)\left(\tan {25}^0\frac{1}{\tan {25}^0}\right)}\\ \end{array} \\ \begin{array}{l}\text{=}\frac{3}{7}-\frac{4}{7}\\ \end{array} \\ \text{=}\frac{3-4}{7}=\frac{-1}{7} \end{gathered}$$