Correct Answer - A
We have, `tan2theta=(2tantheta)/(1-tan^(@)theta)`
Let `theta=(22(1)/(2))rArr2theta=45^(@)`
`thereforetan45^(@)=(2tantheta)/(1-tan^(2)theta)`.
`1=(2tantheta)/(1-tan^(2)theta)`
`rArr1-tan^(2)theta=2tantheta`
`rArrtan^(2)theta+2tantheta-1=0`
`tantheta=(-2pmsqrt(2^(2)-4(1)(-1)))/(2(1))`
`=(-2pmsqrt8)/(2)=-1pmsqrt2`.
As `theta " is " (22(1)/(2)), tan(22(1)/(2))^(@)=sqrt2-1`.
