Question

$\frac{3{\tan }^2{30}^0+{\tan }^2{60}^0+{\csc }^2{30}^0-\tan {45}^0}{{\cot }^2{45}^0}$

Answer 1

$\frac{3{\tan }^2{30}^0+{\tan }^2{60}^0+{\csc }^2{30}^0-\tan {45}^0}{{\cot }^2{45}^0}$

$=\frac{(3\times \frac{1}{3})+3+4-1}{1}$

$=7$

Hence, the answer is $7.$