Question

cot 54° / tan 36° + tan 20°/ cot 70° -2=0​

Answer 1

Step-by-step explanation:

= cot54°/tan(90-54) + tan20°/cot(90-20) - 2

= cot54°/cot54° + tan20°/tan20° - 2

= 1 + 1 - 2

= 0

Answer 2

Answer:

0

Step-by-step explanation:

We have,

$\rm \dfrac{\cot 54^{\circ}}{\tan 36^{\circ}} +\dfrac{\tan 20^{\circ}}{\cot 70^{\circ}}-2$

We know that,

$\rm \tan (90-\theta) = \cot θ$

so,

$\rm \dfrac{\cot 54^{\circ}}{\tan (90-54)} +\dfrac{\tan (90-70)}{\cot 70^{\circ}}-2$

$\rm \dfrac{\cot 54^{\circ}}{\cot 54^{\circ}} +\dfrac{\cot 70^{\circ}}{\cot 70^{\circ}}-2$

$\rm \dfrac{\cancel{\cot 54^{\circ}}}{\cancel{\cot 54^{\circ}}} +\dfrac{\cancel{\cot 70^{\circ}}}{\cancel{\cot 70^{\circ}}}-2$

$\rm 1+1-2$

$\rm 2-2$

$\rm 0$