Question

Solve $\frac{\sec \theta \cos ec({90}^o-\theta )-\tan \theta \cot ({90}^o-\theta )+({\sin }^2{35}^o+{\sin }^2{55}^o)}{\tan {10}^o\tan {20}^o\tan {45}^o\tan {70}^o\tan {80}^o}$

Answer 1

$\frac{\sec \theta cosec\left({90}^0-\theta \right)-\tan \theta \left({90}^0-\theta \right)+\left({\sin }^2{35}^0+{\sin }^2{55}^0\right)}{\tan {10}^0\tan {20}^0\tan {45}^0\tan {70}^0\tan {80}^0}$

$=\frac{\sec \theta \sec \theta -\tan \theta \tan \theta +{\sin }^2{35}^0+{\cos }^2\left({90}^0-{55}^0\right)}{\tan {10}^0\tan {80}^0\tan {20}^0\tan {70}^0\tan {45}^0}$

$=\frac{\left({\sec }^2-{\tan }^2\theta \right)+{\sin }^2{35}^0+{\cos }^2{35}^0}{\left(\tan 10\cot {10}^0\right)\left(\tan {20}^0\cot {20}^0\right)\tan {45}^0}$

$=\frac{1+1}{1\times 1\times 1}$

$=2$