समीकरण `sin^(2) n (theta) - sin^(2) (n-1) theta = sin^(2) theta` से ` theta` का व्यापक मान ज्ञात कीजिए ।

Explanation

` sin^(2) n theta - sin^(2) (n-1) theta = sin^(2) theta `
` rArr sin {n theta - ( n -1) theta} sin { n theta + (n-1) theta} - sin^(2) theta = 0 `
` rArr sin theta * sin ( 2n-1) theta - sin^(2) theta = 0 `
` rArr sin theta [ sin (2n-1) theta - sin theta ] = 0 `
`rArr sin theta * 2 cos n theta * sin ( n-1) theta = 0 `
` sin theta = 0 "या " cos n theta = 0 "या " sin ( n-1) theta = 0 `
यदि `sin theta = 0 , "तो " theta = rpi " जहाँ " r in I . `
यदि ` cos n theta = 0 , "तो " ntheta = (2r+1) pi/2`
` rArr theta = ( 2r +1) pi/(2n)`
यदि ` sin ( n - 1 ) theta =0 , "तो " (n-1) theta = rpi`
` rArr theta = ( rpi)/(n-1)`