Question

Evaluate:
= $\frac{3{\tan }^2{30}^{\circ }+{\tan }^2{60}^{\circ }+\cos ec{30}^{\circ }-\tan {45}^{\circ }}{{\cot }^2{45}^{\circ }}$

Answer 1

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

We know that

$\tan {30}^{\circ }=\frac{1}{\sqrt{3}},\tan {60}^{\circ }=\sqrt{3},\csc {30}^{\circ }=2,\tan {45}^{\circ }=1$ and $\cot {45}^{\circ }=1$

Substituting the above values in

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

$=\frac{3\times {\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\sqrt{3}\right)}^2+2-1}{1}$

$=3\times \frac{1}{3}+3+2-1=1+3+2-1=5$