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sin^2 n theta- sin^2 (n-1)theta= sin^2 t...

`sin^2 n theta- sin^2 (n-1)theta= sin^2 theta` where `n` is constant and `n != 0,1`

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`sin^(2)ntheta-sin^(2)(n-1)theta=sin^(2)theta`
`rArr sin{ntheta-(n-1)theta}sin{ntheta+(n-1)theta}-sin^(2)theta=0`
`rArr sintheta.sin(2n-1)theta-sintheta]=0`
`rArr sintheta.2cosntheta.sin(n-1)theta=0`
`rArr sintheta=0` or `cosntheta=0`
or `sin(n-1)theta=0`
If `sintheta=0`, then `theta=rpi` where `r int I` Ans.
If `cosntheta=0`, then `ntheta=(2r+1)pi/2`
`rArr theta=(2r+1)pi/(2pi)` Ans.
If `sin(n-1)theta=0`, then `(n-1)theta=rpi`
`rArr theta=(rpi)/(n-1)`. Ans.
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