`( sin(n+1) alpha - sin (n-1)alpha )/( cos (n+1) alpha + 2 cos n alpha + cos (n-1) alpha )= `

cos (n + 1) alpha cos (n-1) alpha + sin (n + 1) alpha sin (n-1) alpha =

cos(n+1) alpha * cos(n-1)alpha + sin (n+1)alpha * sin (n-1) alpha =

f_(n)(a)=(sin alpha+sin3 alpha+sin5 alpha+...+sin(2n-1)alpha)/(cos alpha+cos3 alpha+cos5 alpha+...+cos(2n-1)alpha) Then,the value of f_(4)((pi)/(32)) is equal to

1/(cos alpha + cos 3 alpha)+ 1/(cos alpha + cos 5 alpha)+ 1/(cos alpha + cos 7 alpha) +....+ 1/(cos alpha +cos(2n+1) alpha) = 1/2 "cosec" alpha[tan (n+1) alpha - tan alpha ]

if cos^(2)alpha-sin alpha=(1)/(n) then alpha

Using Principle of Mathematical Induction, prove that : cos alpha cos 2 alpha cos 4 alpha ............. cos (2^(n-1) alpha)= (sin (2^n alpha))/(2^n sin alpha) for all n in N.

If n is a positive integer, then the value of (cos alpha cos 2 alpha cos 2 ^(2) alpha ….cos 2 ^(n-1)alpha) is-