Question

$\frac{-\mathrm{tan\theta }·\cot (90°-\mathrm{\theta })+\mathrm{sec\theta }\mathrm{cosec}(90°-\mathrm{\theta })+{\sin }^{2}35°+{\sin }^{2}55°}{\tan 10°\tan 20°\tan 30°\tan 70°\tan 80°}$

Answer 1

$$\begin{gathered} \begin{array}{l}\frac{-\tan \theta \cot ({90}^0-\theta )+\sec \theta \cos \text{ec}({90}^0-\theta )+{\sin }^2{35}^0+{\sin }^2{55}^0}{\tan {10}^0\tan {20}^0\tan {30}^0\tan {70}^0\tan {80}^0}\\ \end{array} \\ \begin{array}{l}=\frac{-\tan \theta \tan \theta +\sec \theta \sec \theta +{\sin }^2{35}^0+{\sin }^2({90}^0-{35}^0)}{\left\{\tan {10}^0\tan ({90}^0-{10}^0)\tan {20}^0\tan ({90}^0-{20}^0)\right\}\tan {30}^0}\\ \end{array} \\ \begin{array}{l}\text{=}\frac{-{\tan }^2\theta +{\sec }^2\theta +({\sin }^2{35}^0+{\cos }^2{35}^0)}{\left(\tan {10}^0\cot {10}^0\right)\left(\tan {20}^0\cot {20}^0\right)\frac{1}{\sqrt{3}}}\\ \end{array} \\ \begin{array}{l}\text{=}\frac{{\sec }^2\theta -{\tan }^2\theta +1}{\left(\tan {10}^0\frac{1}{\tan {10}^0}\right)\left(\tan {20}^0\frac{1}{\tan {20}^0}\right)\frac{1}{\sqrt[]{3}}}\\ \end{array} \\ \begin{array}{l}\text{=}\frac{(1+1)\sqrt{3}}{1}\\ \end{array} \\ =2\sqrt{3} \end{gathered}$$