[tex]\text{Here,}\\\\\text{Opposite side,}~ O = 2\\\\\text{Hypotenuse,}~ H = \sqrt 5 \\\\\text{Adjacent side,}~ A = \sqrt{5-4} = \sqrt 1 =1 ~~ ;[\text{By using Pythagorean theorem}]\\\\\\\sin \theta = \dfrac{O}H = \dfrac 2{\sqrt 5}\\\\\\\cos \theta = \dfrac{A}{H} = \dfrac{1}{\sqrt 5} \\ \\\\ \tan \theta = \dfrac{O}{A} = \dfrac 21 = 2[/tex]
[tex]csc \theta = \dfrac{1}{\sin \theta} = \dfrac{\sqrt 5}2\\\\\\\sec \theta = \dfrac{1}{\cos \theta} = \sqrt 5 \\\\\\\cot \theta = \dfrac 1{\tan \theta} = \dfrac 12[/tex]
