Without using trigonometric tables, prove that:

$\dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2 = 0$

To prove,

$\dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2 = 0$.

Solving, L.H.S. of the equation we get :

$\Rightarrow \dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2 \\[1em] \Rightarrow \dfrac{\text{cot (90° - 36°)}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot (90° - 20°)}} - 2 \\[1em] \Rightarrow \dfrac{\text{tan 36°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{tan 20°}} - 2 \\[1em] \Rightarrow 1 + 1 - 2 \\[1em] \Rightarrow 0.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2 = 0$.