1.

Solve: `sectheta-1=(sqrt(2)-1)tantheta`

Answer» `sectheta-1=(sqrt(2)-1)tantheta`
`rArr (sectheta-1)^(2)=(sqrt(2)-1)^(2).tan^(2)theta`
`=(2+1-2sqrt(2))(sec^(2)theta-1)`
`=(3-2sqrt(2))(sec^(2)theta-1)`
`rArr (sectheta-1)^(2)=3-2sqrt(2))(sectheta-1)(sectheta+1)`
`rArr (3-2sqrt(2)(sectheta-1)(sectheta+1)-(sectheta-1)^(2)=0`
`rArr (sectheta-1)[(3-2sqrt(2))(sectheta+1)-(sectheta-1)]=0`
`rArr (sectheta-1)[(2-2sqrt(2))sectheta+(4-2sqrt(2))]=0`
`rArr (sectheta-1)[(2-2sqrt(2))sectheta-sqrt(2)(-2sqrt(2)+2)]=0`
`rArr (sectheta-1)(2-sqrt(2))(sectheta-sqrt(2))=0`
`rArr sectheta-1=0` or `sectheta-sqrt(2)=0`
Now `sectheta-1=0`
`rArr sectheta=1`
`rArr costheta=1`
`rArr costheta=cos0^(@)`
`rArr theta=2npi+-0=2npi` Ans.
and `sectheta-sqrt(2)=0`
`rArr sectheta=sqrt(2)`
`rArr costheta=1/sqrt(2)`
`rArr costheta=cosi/4`
`rArr theta=2npi+-pi/4`
But the given equation is not satisfied by `theta=2npi-pi/4`, so
`theta=2npi+pi/4`. Ans.


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