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

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

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 =

Let f_n(a)=(sinalpha+sin3alpha+sin5alpha+...+sin(2n-1)alpha)/(cosalpha+cos3alpha+cos5alpha+...+cos(2n-1)alpha) Then, the value of f_4(pi/32) is equal to

Let f_n(a)=(sinalpha+sin3alpha+sin5alpha+...+sin(2n-1)alpha)/(cosalpha+cos3alpha+cos5alpha+...+cos(2n-1)alpha) Then, the value of f_4(pi/32) is equal to

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 tan theta=(sin alpha-cos alpha)/(sin alpha+cos alpha), prove that 2cos^(2)theta=1+sin2 alpha

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