Question

$\frac{\sin 70°}{\cos 20°}+\frac{\mathrm{cosec}36°}{\sec 54°}-\frac{2\cos 43°\mathrm{cosec}47°}{\tan 10°\tan 40°\tan 50°\tan 80°}$

Answer 1

$$\begin{gathered} \begin{array}{l}\frac{\sin {70}^0}{\cos {20}^0}+\frac{\cos \text{ec}{36}^0}{\sec {54}^0}-\frac{2\cos {43}^0\cos \text{ec}{47}^0}{\tan {10}^0\tan {40}^0\tan {50}^0\tan {80}^0}\\ \end{array} \\ \begin{array}{l}\text{=}\frac{\sin ({90}^0-{20}^0)}{\cos {20}^0}+\frac{\cos \text{ec}({90}^0-{54}^0)}{\sec {54}^0}-\frac{2\cos {43}^0\cos \text{ec}({90}^0-{43}^0)}{\tan {10}^0\tan ({90}^0-{10}^0)\tan {40}^0\tan ({90}^0-{40}^0)}\\ \end{array} \\ \begin{array}{l}=\frac{\cos {20}^0}{\cos {20}^0}+\frac{\sec {54}^0}{\sec {54}^0}-\frac{2\cos {43}^0\sec {43}^0}{\tan {10}^0\cot {10}^0\tan {40}^0\cot {40}^0}\\ \end{array} \\ \begin{array}{l}=1+1-\frac{2\cos {43}^0\frac{1}{\cos {43}^0}}{\tan {10}^0\frac{1}{\tan {10}^0}\tan {40}^0\frac{1}{\tan {40}^0}}\\ \end{array} \\ \begin{array}{l}=2-2\\ \end{array} \\ \text{=0} \end{gathered}$$