Question 983893
*[illustration trig].



Using Pythagoras theorem ; {{{(adjacent side)^2}}} = {{{2^2-1^2}}}

Adjacent side = {{{sqrt(3)}}}

sin {{{theta}}} = opp/hypo = 1/2

cos{{{theta}}} = adj/hypo= {{{sqrt(3)/2}}}

tan{{{theta }}} = opp/adj = {{{1/sqrt(3)}}}


cosec, sec and cot are reciprocals of sin cos & tan respectively

cosec {{{theta}}} =2

sec{{{theta}}} = {{{2/sqrt(3)}}}

cot{{{theta}}} = {{{sqrt(3)}}}