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Solve sin^(2)n theta-sin^(2) (n-1) theta...

Solve `sin^(2)n theta-sin^(2) (n-1) theta=sin^(2) theta`.

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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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