If `sqrt(3) tan theta=3 sin theta` and `theta!=0` then find the value of `sin^(2)theta-cos^(2)theta`.

Explanation

`sqrt(3) tan theta=3sin theta`
`:.sqrt(3) (sin theta)/(cos theta)=3 sin theta` ……`[tan theta=(sin theta)/(cos theta)]`
`:.sqrt(3) (sin theta)/(sin theta)=3 cos theta`
`:.3 cos theta=sqrt(3)`
`:.cos theta=(sqrt(3))/3`………… Taking squares of both the sides, we get
`cos^(2) theta=3/9=1/3`...........1
Now `sin^(2) theta-cos^(2) theta`
`=(1-cos^(2) theta)-cos^(2) theta` ........[`sin^(2) theta+cos^(2) theta=1, :. sin ^(2) =1-cos^(2)theta]`
`=1-cos^(2) theta-cos^(2)theta`
`=1-2cos^(2) theta`
`=1-2xx(1/3)`...[From (1)]
`=1-2/3`
`=(3-2)/3=1/3`
The value of `sin^(2) theta-cos^(2) theta` is `1/3`.