Given that `f(ntheta)=(2sin2theta)/(cos2theta-cos4ntheta),` and `f(theta)+f(2theta)+f(3theta)++f(ntheta)=(sinlambdatheta)/(sinthetasinmutheta),` then the value of `mu-lambda` is ______

Answer

1

Explanation

`f(n theta ) = ( 2 sin theta)/( cos 2 theta - cos 4 n theta)`
`= ( 2sin 2theta)/( 2sin ( 2n +1) theta sin (2n-1) theta)`
` = ( sin ((2n+1)theta - (2n-1)theta))/( sin (2n+1) theta sin (2n-1) theta)`
`( sin (2n+1) theta cos (2n -1) theta - cos (2n+1) theta sin (2n -1) theta)/(sin (2n+1) theta sin (2n-1) theta)`
`= cot (2n-1) theta - cot (2n+1) theta `
` therefore f( theta ) + f( 2 theta) + f( 3 theta) + ... = f ( n theta)`
`" " = cot theta - cot ( 2n +1) theta `
`" " = ( sin (2n +1) theta costheta - cos ( 2n +1) theta sin theta)/( sin ( 2n +1) theta sin theta)`
`" " = (sin 2 n theta)/( sin (2n+1) theta sin theta )`