If `sqrt(3)tantheta=1` then find value of `sin^2theta-cos^2theta`

Explanation

`sqrt(3)tantheta=1rArrtantheta=(1)/(sqrt(3))`
Construct `DeltaABC` , right - angled at B such that
`BC=k,AB=sqrt(3k)`
and `ltBAC=theta`
In `DeltaABC`,
From Pythagoras theorem
`AC^(2)=AB^(2)+BC^(2)=(sqrt3k)^(2)+(k)^(2)=4k^(2)`
`rArrAC=2k`
`:. sintheta=(BC)/(AC)=(k)/(2k)=(1)/(2)`
and `costheta=(AB)/(AC)=(sqrt3k)/(2k)=(sqrt(3))/(2)`
Now , `sin^(2)theta-cos^(2)theta=((1)/(2))^(2)-((sqrt(3))/(2))^(2)=(1)/(4)-(3)/(4)=-(1)/(2)`