Correct Answer – a
`sec theta+tan theta=sqrt(3)`………(i)
`rArrsec^(2)theta-tan^(2)theta=1`
`[1+tan^(2)theta=sec^(2)theta]`
`rArr(sec theta-tantheta)(sec theta+tan theta)=1`
`rArr(sec theta-tan theta)(sec theta+tan theta)=1`
`rArr sectheta-tan theta=(1)/sqrt(3)` ……….(ii)
subtract equation (ii) from (i)
`rArr 2tan theta=sqrt(3)-(1)/sqrt(3)`
`rArr 2tan theta=(3-1)/sqrt(3)=(2) /sqrt(3)`
`rArrtan theta=(1)/sqrt(3)=tan 30^(@)`
`rArr theta=30^(@)[tan30^(@)=(1)/sqrt(3)]`
`rArr tanA^(3)theta =tan90^(@)` (underfined)