(sin2A)/(1+cos2 pi)*(cos n)/(1+cos A)

(sin2A)/(1+cos2A)*(cos A)/(1+cos A)=

Prove that (1+sin2A)/(cos2A)=(cos A+sin A)/(cos A-sin A)=tan((pi)/(4)+A)

Prove that : (1+sin 2A)/(cos 2A) = (cos A + sin A)/(cos A - sin A) = tan (pi/4 + A)

(1 + sin2A) / (cos2A) = (cos A + sin A) / (cos A-sin A) = tan ((pi) / (4) +1)

Given, cos 2^(m) theta cos 2^(m + 1) theta….cos 2^(n)theta = (sin 2^(n+1)theta)/(2^(n-m+1)sin 2^(m)theta)"where"2^(m) theta ne kpi, n, m, k in I solve the following "cos" (pi)/(11) "cos" (2pi)/(11) "cos" (3pi)/(11)..."cos" (10pi)/(11) is equal to

If (cos x)/(cos theta)+(sin x)/(sin theta)=(cos y)/(cos theta)+(sin y)/(sin theta)=1 where x-y is not a multiple of 2 pi, then the value of (cos x*cos y)/(cos^(2)theta)+(sin x*sin y)/(sin^(2)theta)=......[theta!=n pi,(2n+1)(pi)/(2),n in z]

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

The value of sin ^(-1)[cos (cos ^(-1)(cos x)+sin ^(-1)(sin x))] where x in((pi)/(2), pi) is

If 2 sin^(2) ((pi//2) cos^(2) x)=1-cos (pi sin 2x), x ne (2n + 1) pi//2, n in I , then cos 2x is equal to

If 2 sin^(2) ((pi//2) cos^(2) x)=1-cos (pi sin 2x), x ne (2n + 1) pi//2, n in I , then cos 2x is equal to