(dn)/(3sin n-4cos n)

I=int(sinn)/((1-cos n)^(2))dn

int (cos n-sin n)/(cos n+sin n)dn

([((cos A+cos B)/(sin A-sin B))^(n)+((sin A+sin B)/(cos A-cos B))^(n) when n is odd,is -

If n is an odd integer than ((cos A + cos B)/( sin A - sin B))^(n) + ((sin A + sin B)/(cos A- cos B))^(n)=

If f_(n)(x) = (sin x)/(cos3x)+(sin 3x)/(cos 3^(2)x) +(sin 3^(2)x)/(cos 3^(3)x) +....+ (sin 3^(n-1)x)/(cos 3^(n)x)"Then" f_(2) ((pi)/(4)) + f_(3) ((pi)/(4))=

(sin (n +1) A +2 sin n A + sin ( n -1) A )/( cos (n +1) A - cos ( n -1) A ) =

(sin (n +1) A + 2 sin n A + sin (n-1) A)/( cos ( n -1) A - cos (n +1) A)=

If A = [(cos theta, sin theta),(-sin theta, cos theta)] then show that for all the positive integers n, A^(n) = [(cos n theta,sin n theta),(-sin n theta,cos n theta)] .

int_(0)^(x)|sin t|dt, where x in(2n pi,(2n+1)pi)n in N, is equal to (A)4n-cos x(B)4n-sin x(C)4n+1-cos x(D)4n-1-cos x

sin (n + 1) A * sin (n-1) A + cos (n + 1) A * cos (n-1) A = cos2A