यहाँ `sec theta=2implies(1)/(cos theta)=2impliescos theta=1/2`
अर्थात `AB =1 ` तथा `AC =2 `
पाइथागोरस प्रमेय से, `BC^(2)=AC^(2)-AB^(2)`
`=(2)^(2)-(1)^(2)=4-1=3`
`impliesBC=sqrt3`
अब `sin theta=(BC)/(AC)=(sqrt3)/(2)`
`impliescosec theta=(1)/(sin theta)=(2)/(sqrt3)`
`tan theta=(BC)/(AB)=(sqrt3)/(1)=sqrt3`
`impliescos theta=(1)/(tan theta)=91)/(sqrt3)`
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