यदि `cosec theta=2,` तो `(1)/(tantheta)+(sintheta)/(1+cos theta)` का मान ज्ञात कीजिए।

Explanation

समकोण `DeltaABC` बनाइए, जिसमे `angle BAC=theta` तथा `cosec theta=2`
किन्तु `cosec theta=(AC)/(BC)`
`thereforeAC=2`और `BC=1`
`DeltaABC` में `angle B=90^(@)`
`AB^(2)+BC^(2)=AC^(2)`
`thereforeAB^(2)+BC^(2)=AC^(2)`
`thereforeAB^(2)+1=4`
`thereforeAB^(2)=4-1`
`thereforeAB^(2)=3`
`thereforeAB=sqrt3`
`thereforetan theta=(BC)/(AC)=(1)/(sqrt3)`
`sin theta=(BC)/(AC)=1/2`
`cos theta=(AB)/(AC)=(sqrt3)/(2)`
`therefore(1)/(tan theta)+(sin theta)/(1+cos theta)=(1)/((1)/(sqrt3))+(1/2)/(1+(sqrt3)/(2))=sqrt3+(1/2)/(+(sqrt3)/(2))=sqrt3+(1/2)/((2+sqrt3)/(2))=sqrt3+(1)/(2+sqrt3)`
`=(sqrt3(2+sqrt3)+1)/(2+sqrt3)=(2sqrt3+3+1)/(2-sqrt3)=(2sqrt3+4)/(2+sqrt3)=(2(sqrt3+2))/(2+sqrt3)=2.`